chapter 11 assessment
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school chemistryTRANSCRIPT
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Solutions Manual Chemistry: Matter and Change Chapter 11 209
StoichiometryStoichiometry
SOLUTIONS MANUALCHAPTER 11
Section 11.1 Defining Stoichiometrypages 368372
Practice Problemspages 371372
1. Interpret the following balanced chemical equa-tions in terms of particles, moles, and mass. Show that the law of conservation of mass is observed.a. N2(g) 3H2(g) 2NH3(g)
1 molecule N2 3 moleculesH2 2 molecules NH3
1 mole N2 3 moles H2 2 moles NH3
Mass:
N2: (2 mol)(14.007 g/mol) 28.014 g
3H2: (6 mol)(1.008 g/mol) 6.048 g
2NH3: (2 mol)(14.007 g/mol) (6 mol)(1.008 g/mol) 34.062 g
28.014 g N2 6.048 g H2 0 34.062 g NH3
34.062 g reactants 34.062 g products
b. HCl(aq) KOH(aq) KCl(aq) H2O(l)1 molecule HCl 1 formula unit KOH 1 formula unit KCl 1 molecule H2O
1 mole HCl 1 mole KOH 1 mole KCl 1 mole H2O
Mass:
HCl: (1 mol)(1.008 g/mol) (1 mol)(35.453 g/mol) 36.461 g
KOH: (1 mol)(39.098 g/mol) (1 mol)(15.999 g/mol) (1 mol)(1.008 g/mol) 56.105 g
KCl: (1 mol)(39.098 g/mol) (1 mol)(35.453 g/mol) 74.551 g
H2O: (2 mol)(1.008 g/mol) (1 mol)( 15.999 g/mol) 18.015 g
36.461 g HCl 56.105 g KOH 74.551 g KCl 18.015 g H2O
92.566 g reactants 92.566 g products
c. 2Mg(s) O2(g) 2MgO(s)2 atoms Mg 1 moleculeO2 0 2 formula units MgO
2 moles Mg 1 mole O2 2 moles MgO
Mass:
2Mg: (2 mol)(24.305 g/mol) 48.610 g
O2: (2 mol)(15.999 g/mol) 31.998 g
2MgO: (2 mol)(24.305 g/mol) (2 mol)(15.999 g/mol) 80.608 g
48.610 g Mg 31.998 g O2 80.608 g MgO
80.608 g reactants 80.608 g products
2. Challenge For each of the following, balance the chemical equation; interpret the equation in terms of particles, moles, and mass; and show that the law of conservation of mass is observed.a. ___Na(s) ___H2O(l) ___NaOH(aq)
___H2(g)2Na(s) 2H2O(l) 2NaOH(aq) 1H2(g)
2 atoms Na 2 molecules H2O 2 formula units NaOH 1 molecule H2
2 mol Na 2 mol H2O 2 mol NaOH 1 mol H2
Mass:
2Na: (2 mol)(22.990 g/mol) 45.980 g
2H2O: (4 mol)(1.008 g/mol) (2 mol)(15.999 g/mol) 36.030 g
2NaOH: (2 mol)(22.990 g/mol) (2 mol)(15.999 g/mol) (2 mol)(1.008 g/mol) 79.994 g
H2: (2 mol)(1.008 g/mol) 2.016 g
45.980 g Na 36.030 g H2O 79.994 g NaOH 2.016 g H2
82.01 g reactants 82.01 g products
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210 Chemistry: Matter and Change Chapter 11 Solutions Manual
SOLUTIONS MANUALCHAPTER 11
b. ___Zn(s) ___HNO3(aq) ___Zn(NO3)2(aq) ___N2O(g) ___H2O(l)4Zn(s) 10HNO3(aq) 4Zn(NO3)2(aq) 1N2O(g) 5H2O(l)
4 atoms Zn 10 molecules HNO3 4 formula units Zn(NO3)2 1 molecule N2O 5 molecules H2O
4 mol Zn 10 mol HNO3 4 mol Zn(NO3)2 1 mol N2O 5 mol H2O
Mass:
4Zn: (4 mol)(65.39 g/mol) 261.56 g
10HNO3: (10 mol)(1.008 g/mol) (10 mol)(14.007 g/mol) (30 mol)(15.999 g/mol) 630.12 g
4Zn(NO3)2: (4 mol)(65.39 g/mol) (8 mol)(14.007 g/mol) (24 mol)(15.999 g/mol) 757.592 g
N2O: (2 mol)(14.007 g/mol) (1 mol)(15.999 g/mol) 44.013 g
5H2O: (10 mol)(1.008 g/mol) (5 mol)(15.999 g/mol) 90.075 g
261.56 g Zn 630.12 g HNO3 757.592 g Zn(NO3)2 44.013 g N2O 90.075 g H2O
891.68 g reactants 891.68 g products
3. Determine all possible mole ratios for the following balanced chemical equations.a. 4Al(s) 3O2(g) 2Al2O3(s)
4 mol Al _ 3 mol O2
3 mol O2 __
2 mol Al2O3
2 mol Al2O3 __ 4 mol Al
3 mol O2 _ 4 mol Al
2 mol Al2O3 __
3 mol O2 4 mol Al __
2 mol Al2O3
b. 3Fe(s) 4H2O(l) Fe3O4(s) 4H2(g) 3 mol Fe __ 4 mol H2O
3 mol Fe _ 4 mol H2
3 mol Fe __ 1 mol Fe3O4
4 mol H2O __
3 mol Fe
4 mol H2 _ 3 mol Fe
1 mol Fe3O4 __
3 mol Fe
1 mol Fe3O4 __
4 mol H2
1 mol Fe3O4 __ 4 mol H2O
4 mol H2O __
4 mol H2
4 mol H2 __
1 mol Fe3O4
4 mol H2O __ 1 mol Fe3O4
4 mol H2 __ 4 mol H2O
c. 2HgO(s) 2Hg(l) O2(g)
2 mol HgO
__ 2 mol Hg
1 mol O2 _ 2 mol Hg
1 mol O2 __ 2 mol HgO
2 mol Hg
__ 2 mol HgO
2 mol Hg
_ 1 mol O2
2 mol HgO
__ 1 mol O2
4. Challenge Balance the following equations, and determine the possible mole ratios.a. ZnO(s) HCl(aq) ZnCl2(aq) H2O(l)
ZnO(s) 2HCl(aq) ZnCl2(aq) H2O(l)
1 mol ZnO __ 2 mol HCl
1 mol ZnO __ 1 mol ZnCl2
1 mol ZnO __ 1 mol H2O
2 mol HCl __ 1 mol ZnO
2 mol HCl __ 1 mol ZnCl2
2 mol HCl __ 1 mol H2O
1 mol ZnCl2 __ 1 mol ZnO
1 mol ZnCl2 __
2 mol HCl
1 mol ZnCl2 __ 1 mol H2O
1 mol H2O __ 1 mol ZnO
1 mol H2O __ 2 mol HCl
1 mol H2O __ 1 mol ZnCl2
b. butane (C4H10) oxygen carbon dioxide water2C4H10(g) 13O2(g) 0 8CO2(g) 10H2O(l)
2 mol C4H10 __ 13 mol O2
2 mol C4H10 __ 8 mol CO2
2 mol C4H10 __ 10 mol H2O
13 mol O2 __
2 mol C4H10
8 mol CO2 __ 2 mol C4H10
10 mol H2O __ 2 mol C4H10
10 mol H2O __ 13 mol O2
10 mol H2O __ 8 mol CO2
8 mol CO2 _ 13 mol O2
13 mol O2 __
10 mol H2O
8 mol CO2 __ 10 mol H2O
13 mol O2 _ 8 mol CO2
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Solutions Manual Chemistry: Matter and Change Chapter 11 211
SOLUTIONS MANUALCHAPTER 11
Section 11.1 Assessmentpage 372
5. Compare the mass of the reactants and the mass of the products in a chemical reaction, and explain how these masses are related.The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products. The masses of reactants and products are equal.
6. State how many mole ratios can be written for a chemical reaction involving three substances.n 3, thus (n)(n 1) (3)(2) 6 mole ratios
7. Categorize the ways in which a balancedchemical equation can be interpreted.particles (atoms, molecules, formula units), moles, and mass
8. Apply The general form of a chemical reac-tion is xA yB zAB. In the equation, A and B are elements and x, y, and z are coefficients. State the mole ratios for this reaction.xA/yB and xA/zAB
yB/xA and yB/zAB
zAB/xA and zAB/yB
9. Apply Hydrogen peroxide (H2O2) decomposes to produce water and oxygen. Write a balanced chemical equation for this reaction, and deter-mine the possible mole ratios.2H2O2 2H2O O2
2 mol H2O2/2 mol H2O, 2 mol H2O2/1 mol O2, 2 mol H2O/2 mol H2O2, 2 mol H2O/1 mol O2, 1 mol O2/2 mol H2O2, 1 mol O2/2 mol H2O
10. Model Write the mole ratios for the reaction of hydrogen gas and oxygen gas, 2H2(g) O2(g) 2H2O. Make a sketch of six hydrogen molecules reacting with the correct number of oxygen molecules. Show the water molecules produced.
6H2 3O2 6H2O
2H2/O2 and 2H2/2H2O
O2/2H2 and O2/2H2O
2H2O/2H2 and 2H2O/O2
Sketches should show six hydrogen atoms combining with three oxygen atoms to form six water molecules.
Section 11.2 Stoichiometric Calculationspages 373378
Practice Problemspages 375377
11. Methane and sulfur react to produce carbon disulfide (CS2), a liquid often used in the production of cellophane.
___CH4(g) ___S8(s) ___CS2(l) ___H2S(g)a. Balance the equation.
2CH4(g) S8(s) 2CS2(l) 4H2S(g)
b. Calculate the moles of CS2 produced when 1.50 moles of S8 is used.
1.50 mol S8 2 mol CS2 _ 1 mol S8
3.00 mol CS2
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212 Chemistry: Matter and Change Chapter 11 Solutions Manual
SOLUTIONS MANUALCHAPTER 11
c. How many moles of H2S are produced?
1.50 mol S8 4 mol H2S _ 1 mol S8
6.00 mol H2S
12. Challenge Sulfuric acid (H2SO4) is formed when sulfur dioxide (SO2) reacts with oxygen and water.a. Write the balanced chemical equation for the
reaction.
2SO2(g) O2(g) 2H2O(l) 2H2SO4(aq)
b. How many moles of H2SO4 are produced from 12.5 moles of SO2?
12.5 mol SO2 2 mol H2SO4 __
2 mol SO2
12.5 mol H2SO4 produced
c. How many moles of O2 are needed?
12.5 mol SO2 1 mol O2 _ 2 mol SO2
6.25 mol O2 needed
13. Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy, as shown below. How much chlorine gas, in grams, is obtained from the process?
2.50 mol ? g
Electricenergy NaCl Cl2
Na
Step 1: Balance the chemical equation.
2NaCl(s) 2Na(s) Cl2(g)
Step 2: Make mole mole conversion.
2.50 mol NaCl 1 mol Cl2 __
2 mol NaCl 1.25 mol Cl2
Step 3: Make mole mass conversion.
1.25 mol Cl2 70.9 g Cl2 _ 1 mol Cl2
88.6 g Cl2
14. Challenge Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4) is extracted from titanium oxide (TiO2) using chlorine and coke (carbon).
TiO2(s) C(s) 2Cl2(g) TiCl4(s) CO2(g)a. What mass of Cl2 gas is needed to react with
1.25 mol of TiO2?Step 1: Make mole mole conversion.
1.25 mol TiO2 2 mol Cl2 __
1 mol TiO2 2.50 mol Cl2
Step 2: Make mole mass conversion.
2.50 mol Cl2 70.90 g Cl2 __ 1 mol Cl2
177 g Cl2
b. What mass of C is needed to react with 1.25 mol of TiO2?Step 1: Make mole mole conversion.
1.25 mol TiO2 1 mol C __
1 mol TiO2 1.25 mol C
Step 2: Make mole mass conversion.
1.25 mol C 12.011 g C
__ 1 mol C
15.0 g C
c. What is the mass of all of the products formed by reaction with 1.25 mol of TiO2?Calculate the mass to TiO2 used.
1.25 mol TiO2 79.865 g TiO2 __
1 mol TiO2 99.8 g TiO2
Calculate the total mass of the reactants.
99.8 g 177 g 15.0 g 292 g
Because mass is conserved, the mass of the products must equal the mass of reactants.
mass of products 292 g
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Solutions Manual Chemistry: Matter and Change Chapter 11 213
SOLUTIONS MANUALCHAPTER 11
15. One of the reactions used to inflateautomobile air bags involves sodium azide (NaN3): 2NaN3(s) 2Na(s) 3N2(g). Determine the mass of N2 produced from the decomposition of NaN3 shown below.
100.0 g NaN3 ? g N2(g)
N2 gas
2NaN3(s) 2Na(s) 3N2(g)
Step 1: Make mass mole conversion.
100.0 g NaN3 1 mol NaN3 __
65.02 g NaN3 1.538 mol NaN3
Step 2: Make mole mole conversion.
1.538 mol NaN3 3 mol N2 __
2 mol NaN3 2.307 mol N2
Step 3: Make mole mass conversion.
2.307 mol N2 28.02 g N2 __ 1 mol N2
64.64 g N2
16. Challenge In the formation of acid rain, sulfur dioxide (SO2) reacts with oxygen and water in the air to form sulfuric acid (H2SO4). Write the balanced chemical equation for the reac-tion. If 2.50 g of SO2 reacts with excess oxygen and water, how much H2SO4, in grams, is produced?Step 1: Balance the chemical equation.
2SO2(g) O2(g) 2H2O(l) 0 2H2SO4(aq)
Step 2: Make mass mole conversion.
2.50 g SO2 1 mol SO2 __
64.07 g SO2 0.0390 mol SO2
Step 3: Make mole mole conversion.
0.0390 mol SO2 2 mol H2SO4 __
2 mol SO2
0.0390 mol H2SO4
Step 4: Make mole mass conversion.
0.0390 mol H2SO4 98.09 g H2SO4 __ 1 mol H2SO4
3.83 g H2SO4
Section 11.2 Assessmentpage 378
17. Explain why a balanced chemical equation is needed to solve a stoichiometric problem.The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products.
18. List the four steps used in solving stoichio-metric problems.1. Balance the equation.
2. Convert the mass of the known substance to moles of known substance.
3. Use the mole ratio to convert from moles of the known to moles of the unknown.
4. Convert moles of unknown to mass of the unknown.
19. Describe how a mole ratio is correctly expressed when it is used to solve a stoichio-metric problem.moles of unknown/moles of known
20. Apply How can you determine the mass of liquid bromine (Br2) needed to react completely with a given mass of magnesium?Write a balanced equation. Convert the given mass of magnesium to moles. Use the mole ratio from the balanced equation to convert moles of magnesium to moles of bromine. Convert from moles of bromine to mass of bromine.
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214 Chemistry: Matter and Change Chapter 11 Solutions Manual
SOLUTIONS MANUALCHAPTER 11
21. Calculate Hydrogen reacts with excess nitrogen as follows:
N2(g) 3H2(g) 2NH3(g) If 2.70 g of H2 reacts, how many grams of NH3
is formed?
(2.70 g H2) ( 1 mol __ 2.016 g H2 ) 1.34 mol H21.34 mol H2 ( 2 mol NH3 _ 3 mol H2 ) 0.893 mol NH3(0.893 mol NH3) ( 17.031 g NH3 __ 1 mol NH3 ) 15.2 g NH3
22. Design a concept map for the following reaction.
CaCO3(s) 2HCl(aq) CaCl2(aq) H2O(l) CO2(g)
The concept map should explain how to deter-mine the mass of CaCl2 produced from a given mass of HCl.Concept maps will vary, but all should show the use of these conversion factors: the inverse of molar mass, the mole ratio, the molar mass.
Section 11.3 Limiting Reactantspages 379384
Practice Problemspage 383
23. The reaction between solid sodium and iron(III) oxide is one in a series of reactions that inflates an automobile airbag: 6Na(s) Fe2O3(s) 3Na2O(s) 2Fe(s). If 100.0 g of Na and 100.0 g of Fe2O3 are used in this reaction, determine the following.a. limiting reactant
Make mass mole conversion.
100.0 g Na 1 mol Na __ 22.99 g Na
4.350 mol Na
100.0 g Fe2O3 1 mol Fe2O3 __
159.7 g Fe2O3
0.6261 mol Fe2O3
Make mole ratio comparison.
0.6261 mol Fe2O3 __
4.350 mol Na compared to
1 mol Fe2O3 __ 6 mol Na
0.1439 compared to 0.1667
The actual ratio is less than the needed ratio, so iron(III) oxide is the limiting reactant.
b. reactant in excessSodium is the excess reactant.
c. mass of solid iron producedMake mole mole conversion.
0.6261 mol Fe2O3 2 mol Fe __
1 mol Fe2O3
1.252 mol Fe
Make mole mass conversion.
1.252 mol Fe 55.85 g Fe
_ 1 mol Fe
69.92 g Fe
d. mass of excess reactant that remains after the reaction is completeMake mole mole conversion.
0.6261 mol Fe2O3 6 mol Na __
1 mol Fe2O3
3.757 mol Na needed
Make mole mass conversion.
3.757 mol Na 22.9 g Na
_ 1 mol Na
86.37 g Na needed
100.0 g Na given 86.37 g Na needed 13.6 g Na in excess
24. Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis.a. Write the balanced chemical equation for the
reaction.6CO2(g) 6H2O(l) 0 C6H12O6(aq) 6O2(g)
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Solutions Manual Chemistry: Matter and Change Chapter 11 215
SOLUTIONS MANUALCHAPTER 11
b. Determine the limiting reactant.Make mass mole conversion.
88.0 g CO2 1 mol CO2 __
44.01 g CO2 2.00 mol CO2
64.0 g H2O 1 mol H2O __
18.02 g H2O 3.55 mol H2O
Make mole ratio comparison.
2.00 mol CO2 __ 3.55 mol H2O
compared to 6 mol CO2 __ 6 mol H2O
;
0.563 compared to 1.00
The actual ratio is less than the needed ratio, so carbon dioxide is the limiting reactant.
c. Determine the excess reactant.Water is the excess reactant.
d. Determine the mass in excess.Make mole mole conversion.
2.00 mol CO2 6 mol H2O __ 6 mol CO2
2.00 mol H2O
Make mole mass conversion.
2.00 mol H2O 18.02 g H2O __ 1 mol H2O
36.0 g H2O needed
64.0 g H2O given 36.0 g H2O needed 28.0 g H2O in excess
e. Determine the mass of glucose produced.Make mole mole conversion.
2.00 mol CO2 1 mol C6H12O6 __
6 mol CO2
0.333 mol C6H12O6Make mole 0 mass conversion.
0.333 mol C6H12O6 180.24 g C6H12O6 __
1 mol C6H12O6
60.0 g C6H12O6
Section 11.3 Assessmentpage 384
25. Describe the reason why a reaction between two elements comes to an end.one of the reactants is used up
26. Identify the limiting and the excess reactant in each reaction.a. Wood burns in a campfire.
The wood limits. Oxygen is in excess. The fire will burn only while wood is present.
b. Airborne sulfur reacts with the silver plating on a teapot to produce tarnish (silver sulfide).Silver is the limiting reactant. Sulfur is in excess. When a layer of tarnish covers the silver surface, it prevents the sulfur in the air from reacting.
c. Baking powder in batter decomposes to produce carbon dioxide.A decomposition reaction usually has only one reactant. The reaction is limited by the amount of baking powder present.
27. Analyze Tetraphosphorous trisulphide (P4S3) is used in the match heads of some matches. It is produced in the reaction 8P4 3S8 8P4S3. Determine which of the following statements are incorrect, and rewrite the incorrect state-ments to make them correct.a. 4 mol P4 reacts with 1.5 mol S8 to form
4 mol P4S3.correct
b. Sulfur is the limiting reactant when 4 mol P4 and 4 mol S8 react.Phosphorus is the limiting reactant.
c. 6 mol P4 reacts with 6 mol S8, forming 1320 g P4S3.correct
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216 Chemistry: Matter and Change Chapter 11 Solutions Manual
SOLUTIONS MANUALCHAPTER 11
Section 11.4 Percent Yieldpages 385388
Practice Problemspage 387
28. Aluminum hydroxide (Al(OH)3) is often present in antacids to neutralize stomach acid (HCl). The reaction occurs as follows: Al(OH)3(s) 3HCl(aq) AlCl3(aq) 3H2O(l). If 14.0 g of Al(OH)3 is present in an antacid tablet, deter-mine the theoretical yield of AlCl3 produced when the tablet reacts with HCl.Make mass mole conversion.
14.0 g Al(OH)3 1 mol Al(OH)3 __ 78.0 g Al(OH)3
0.179 mol Al(OH)3
Make mole mole conversion.
0.179 mol Al(OH)3 1 mol AlCl3 __
1 mol Al(OH)3
0.179 mol AlCl3
Make mole mass conversion.
0.179 mol AlCl3 133.3 g AlCl3 __ 1 mol AlCl3
23.9 g AlCl3
23.9 g of AlCl3 is the theoretical yield.
29. Zinc reacts with iodine in a synthesis reaction:
Zn I2 0 ZnI2.a. Determine the theoretical yield if 1.912 mol
of zinc is used.Write the balanced chemical equation.
Zn(s) I2(s) ZnI2(s)
Make mole mole conversion.
1.912 mol Zn 1 mol ZnI2 _ 1 mol Zn
1.912 mol ZnI2
Make mole mass conversion.
1.912 mol ZnI2 319.2 g ZnI2 __ 1 mol ZnI2
610.3 g ZnI2
610.3 g of ZnI2 is the theoretical yield.
b. Determine the percent yield if 515.6 g of product is recovered.
% yield 515.6 g ZnI2 __ 610.3 g ZnI2
100
84.48% yield of ZnI2
30. Challenge When copper wire is placed into a silver nitrate solution (AgNO3), silver crystals and copper(II) nitrate (Cu(NO3)2) solution form.a. Write the balanced chemical equation for the
reaction.Cu(s) 2AgNO3(aq) 2Ag(s) Cu(NO3)2(aq)
b. If a 20.0g sample of copper is used, determine the theoretical yield of silver.Make mass mole conversion.
20.0 g Cu 1 mol Cu __ 63.55 g Cu
0.315 mol Cu
Make mole mole conversion.
0.315 mol Cu 2 mol Ag
_ 1 mol Cu
0.630 mol Ag
Make mole mass conversion.
0.630 mol Ag 107.9 g Ag
__ 1 mol Ag
68.0 g Ag
68.0 g of Ag is the theoretical yield.
c. If 60.0 g of silver is recovered from the reaction, determine the percent yield of the reaction.
% yield 60.0 g Ag
_ 68.0 g Ag
100
88.2% yield of Ag
Section 11.4 Assessmentpage 388
31. Identify which type of yieldtheoretical yield, actual yield, or percent yieldis a measure of the efficiency of a chemical reaction.percent yield
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Solutions Manual Chemistry: Matter and Change Chapter 11 217
SOLUTIONS MANUALCHAPTER 11
32. List several reasons why the actual yield from a chemical reaction is not usually equal to the theoretical yield.Not all reactions go to completion. Some of the reactants or products stick to the surface of the container and are not massed or transferred. Other unexpected products form from competing reactions.
33. Explain how percent yield is calculated.divide the actual yield by the theoretical yield and multiply the quotient by 100
34. Apply In an experiment, you combine 83.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(III) sulfide.
2Fe(s) 3S(s) Fe2S3(s) What is the theoretical yield, in grams, of
iron(III) sulfide?(83.77 g Fe) ( 1 mol Fe __ 55.845 g Fe ) 1.500 mol Fe1.500 mol Fe ( 1 mol Fe2S3 __ 2 mol Fe ) ( 207.885 g Fe2S3 __ 1 mol Fe2S3 ) 155.9 g Fe2S3
35. Calculate the percent yield of the reaction of magnesium with excess oxygen:
2Mg(s) O2(g) 2MgO(s)Reaction Data
Mass of empty crucible 35.67 g
Mass of crucible and Mg 38.06 g
Mass of crucible and MgO (after heating)
39.15 g
mass of Mg 38.06 g 35.67 g 2.39 g
mass of MgO 39.15 g 33.67 g 3.48 g
Theoretical yield: 2.39 g Mg 1 mol Mg
__ 24.31 g Mg
0.0983 mol Mg
0.983 mol Mg 2 mol MgO
__ 2 mol Mg
40.31 g MgO
__ 1 mol MgO
3.96 g MgO
% yield 3.48 g MgO
__ 3.96 g MgO
100
87.9% yield of MgO
Chapter 11 Assessmentpages 392397
Section 11.1
Mastering Concepts 36. Why must a chemical equation be balanced
before you can determine mole ratios?Mole ratios are determined by the coefficients in a balanced equation. If the equation is not balanced, the relationship between reactants and products cannot be determined.
37. What relationships can be determined from a balanced chemical equation?The relationships among particles, moles, and mass for all reactants and products.
38. Explain why mole ratios are central to stoichiometric calculations.Mole ratios allow for the conversion from moles of one substance in a balanced chemical equation to moles of another substance in the same equation.
39. What is the mole ratio that can convert from moles of A to moles of B?
moles B _ moles A
40. Why are coefficients used in mole ratios instead of subscripts?The coefficients in the balanced chemical equation show the numbers of representative particles involved in a reaction. Subscripts give the numbers of different kinds of atoms within a molecule or formula unit.
41. Explain how the conservation of mass allows you to interpret a balanced chemical equation in terms of mass.The mass of the reactants will always equal the mass of the products.
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218 Chemistry: Matter and Change Chapter 11 Solutions Manual
SOLUTIONS MANUALCHAPTER 11
42. When heated by a flame, ammonium dichro-mate decomposes, producing nitrogen gas, solid chromium(III) oxide, and water vapor.
(NH4)2Cr2O7 N2 Cr2O3 4H2O Write the mole ratios for this reaction that relate
ammonium chromate to the products.
1 mol (NH4)2Cr2O7 __
1 mol N2
1 mol (NH4)2Cr2O7 __ 1 mol Cr2O3
1 mol (NH4)2Cr2O7 __
4 mol H2O
1 mol N2 __
1 mol (NH4)2Cr2O7
1 mol Cr2O3 __ 1 mol (NH4)2Cr2O7
4 mol H2O __
1 mol (NH4)2Cr2O7
43. Figure 11.10 depicts an equation with squares representing Element M and circles repre-senting Element N. Write a balanced equation to represent the picture shown, using smallest whole-number ratios. Write mole ratios for this equation.2M2N M4 N2
2 mol M2N __ 1 mol M4
, 2 mol M2N __ 1 mol N2
,
1 mol M4 _ 1 mol N2
, 1 mol M4 __
2 mol M2N ,
1 mol N2 __
2 mol M2N ,
1 mol N2 _ 1 mol M4
Mastering Problems 44. Interpret the following equation in terms of
particles, moles, and mass.
4Al(s) 3O2(g) 2Al2O3(s)Particles: 4 atoms Al 3 molecules O2 2 formula units Al2O3
Moles: 4 mol Al 3 mol O2 2 mol Al2O3;
Mass:
4Al 4 mol Al(26.982 g/mol) 107.93 g
3O2 6 mol O(15.999 g/mol) 95.99 g
2Al2O3 4 mol Al(26.982 g/mol) 6 mol O(15.999 g/mol) 203.92 g
107.93 g Al 95.99 g O2 203.92 g Al2O3
45. Smelting When tin(IV) oxide is heated with carbon in a process called smelting, the element tin can be extracted.
SnO2(s) 2C(s) Sn(l) 2CO(g) Interpret the chemical equation in terms of
particles, moles, and mass.Particles: 1 formula unit SnO2 2 atoms C 1 atom Sn 2 molecules CO
Moles: 1 mol SnO2 2 molC 1 mol Sn 2 mol CO
Mass:
SnO2: 1 mol(118.710 g/mol) 2 mol (15.999 g/mol) 150.71 g
2C: 2 mol(12.011 g/mol) 24.02 g
Sn: 1 mol(118.710 g/mol) 118.71 g
2CO: 2 mol(12.011 g/mol) 2 mol(15.999 g/mol) 56.02 g
150.71 g SnO2 24.02 g C 118.71 g Sn 56.02 g CO
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46. When solid copper is added to nitric acid, copper(II) nitrate, nitrogen dioxide, and water are produced. Write the balanced chemical equation for the reaction. List six mole ratios for the reaction.Cu(s) 4HNO3(aq) Cu(NO3)2(aq) 2NO2(g) 2H2O(l); answers should include any six of the following ratios: 1 mol Cu/4 mol HNO3, 1 mol Cu/1 mol Cu(NO3)2, 1 mol Cu/2 mol NO2, 1 mol Cu/2 mol H2O, 4 mol HNO3/1 mol Cu, 4 mol HNO3/1 mol Cu(NO3)2, 4 mol HNO3/2 mol NO2, 4 mol HNO3/2 mol H2O, 1 mol Cu(NO3)2/1 mol Cu, 1 mol Cu(NO3)2/4 mol HNO3, 1 mol Cu(NO3)2/2 mol NO2, 1 mol Cu(NO3)2/2 mol H2O, 2 mol NO2/1 mol Cu, 2 mol NO2/4 mol HNO3, 2 mol NO2/1 mol Cu(NO3)2, 2 mol NO2/2 mol H2O, 2 mol H2O/1 mol Cu, 2 mol H2O/4 mol HNO3, 2 mol H2O/1 mol Cu(NO3)2, 2 mol H2O/2 mol NO2
47. When hydrochloric acid solution reacts with lead(II) nitrate solution, lead(II) chloride precip-itates and a solution of nitric acid is produced.a. Write the balanced chemical equation for the
reaction.2HCl(aq) Pb(NO3)2(aq) PbCl2(s) 2HNO3(aq)
b. Interpret the equation in terms of molecules and formula units, moles, and mass.Particles: 2 molecules HCl 1 formula unit Pb(NO3)2 1 formula unit PbCl2 2 molecules HNO3
Moles: 2 mol HCl 1 mol Pb(NO3)2 1 mol PbCl2 2 mol HNO3;
Mass:
2HCl: 2 mol(1.008 g/mol) 2 mol (35.453 g/mol) 72.9 g
Pb(NO3)2: 1 mol(207.2 g/mol) 2 mol (14.007 g/mol) 6 mol(15.999 g/mol) 331.2 g
PbCl2: 1 mol(207.2 g/mol) 2 mol(35.453 g/mol) 278.1 g
2HNO3: 2 mol(1.008 g/mol) 2 mol(14.007 g/mol) 6 mol(15.999 g/mol) 126.0 g
72.9 g HCl 331.2 g Pb(NO3)2 278.1 g PbCl2 126.0 g HNO3
48. When aluminum is mixed with iron(III) oxide, iron metal and aluminum oxide are produced along with a large quantity of heat. What mole ratio would you use to determine moles of Fe if moles of Fe2O3 are known?
Fe2O3(s) 2Al(s) 2Fe(s) Al2O3(s) heat 2 mol Fe __ 1 mol Fe2O3
49. Solid silicon dioxide, often called silica, reacts with hydrofluoric acid (HF) solution to produce the gas silicon tetrafluoride and water.a. Write the balanced chemical equation for the
reaction.SiO2(s) 4HF(aq) SiF4(g) 2H2O(l)
b. List three mole ratios, and explain how you would use them in stoichiometric calculations.Students may write any of 12 ratios. Examples might be the following. 4 mol HF/1 mol SiO2; used to find the amount of HF that will react with a known amount of SiO2; 1 mol SiF4/1 mol SiO2; used to find the amount of SiF4 that can be formed from a known amount of SiO2. 2 mol H2O/1 mol SiF4; used to find the amount of H2O that will be produced with the SiF4
50. Chrome The most important commercial ore of chromium is chromite (FeCr2O4). One of the steps in the process used to extract chromium from the ore is the reaction of chromite with coke (carbon) to produce ferrochrome (FeCr2).
2C(s) FeCr2O4(s) FeCr2(s) 2CO2(g) What mole ratio would you use to convert from
moles of chromite to moles of ferrochrome?
1 mol FeCr2 __
1 mol FeCr2O4
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51. Air Pollution The pollutant SO2 is removed from the air in a reaction that also involves calcium carbonate and oxygen. The products of this reaction are calcium sulfate and carbon dioxide. Determine the mole ratio you would use to convert moles of SO2 to moles of CaSO4.2SO2 2CaCO3 O2 2CaSO4 2CO2
2 mol CaSO4 __
2 mol SO2
52. Two substances, W and X, react to form the products Y and Z. Table 11.2 shows the moles of the reactants and products involved when the reaction was carried out. Use the data to determine the coefficients that will balance the equation W X Y Z.
Reaction Data
Moles of Reactants Moles of Products
W X Y Z
0.90 0.30 0.60 1.20
Divide each molar quantity by 0.30 mol, the least common denominator:
W: 0.90/0.30 3
X: 0.30/0.30 1
Y: 0.60/0.30 2
Z: 1.20/0.30 4
3W X 0 2Y 4Z
53. Antacids Magnesium hydroxide is an ingre-dient in some antacids. Antacids react with excess hydrochloric acid in the stomach to relieve indigestion.
___Mg(OH)2 ___HCl ___ MgCl2 ___ H2Oa. Balance the reaction of Mg(OH)2 with HCl.
1Mg(OH)2 2HCl 1MgCl2 2H2O
b. Write the mole ratio that would be used to determine the number of moles of MgCl2 produced when HCl reacts with Mg(OH)2.
1 mol MgCl2 __
1 mol Mg(OH)2 or
1 mol MgCl2 __ 2 mol HCl
Section 11.2
Mastering Concepts 54. What is the first step in all stoichiometric
calculations?Write a balanced chemical equation for the reaction.
55. What information does a balanced equation provide?The balanced equation provides the relationship between reactants and products, and the coefficients in the equation are used to write mole ratios relating reactants and products.
56. On what law is stoichiometry based, and how do the calculations support this law?Stoichiometry is based on the law of conservation of mass. The calculations are used to determine the mass of reactants and products. Once found, the sum of reactants will equal the sum of products, verifying the law of conservation of mass.
57. How is molar mass used in some stoichiometric calculations?Molar mass is a conversion factor for converting moles of a given substance to mass or mass of a given substance to moles.
58. What information must you have in order to calculate the mass of product formed in a chemical reaction?You must have the balanced chemical equation and know the quantity of one substance in the reaction other than the product you are to determine.
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+
59. Each box in Figure 11.11 represents the contents of a flask. One flask contains hydrogen sulfide, and the other contains oxygen. When the contents of the flasks are mixed, a reaction occurs and water vapor and sulfur are produced. In the figure, the red circles represent oxygen, the yellow circles represent sulfur, and blue circles represent hydrogen.a. Write the balanced chemical equation for the
reaction.2H2S(g) O2(g) 2H2O(g) 2S(s)
b. Using the same color code, sketch a representation of the flask after the reaction occurs.
Student sketches should show the formation of six water molecules (H2O) and six sulfur atoms (S).
Mastering Problems 60. Ethanol (C2H5OH ), also known as grain
alcohol, can be made from the fermentation of sugar (C6H12O6). The unbalanced chemical equation for the reaction is shown below.
___C6H12O6 ___C2H5OH ___CO2 Balance the chemical equation and determine
the mass of C2H5OH produced from 750 g of C6H12O6.C6H12O6 2C2H5OH 2CO2
750 g C6H12O6(180.16 g/mol) 4.2 mol C6H12O6
4.2 mol C6H12O6 ( 2 mol C2H5OH __ 1 mol C6H12O6 ) 8.4 mol C2H5OH
8.4 mol C2H5OH ( 46.07 g __ 1 mol C2H5OH ) 390 g C2H5OH
61. Welding If 5.50 mol of calcium carbide (CaC2) reacts with an excess of water, how many moles of acetylene (C2H2), a gas used in welding, will be produced?
CaC2(s) 2H2O(l) Ca(OH)2(aq) C2H2(g)The mole ratio of CaC2:C2H2 is 1:1; thus, 5.50 mol C2H2 will be produced from 5.50 mol CaC2.
62. Antacid Fizz When an antacid tablet dissolves in water, the fizz is due to a reaction between sodium hydrogen carbonate (NaHCO3), also called sodium bicarbonate, and citric acid (H3C6H5O7).
3NaHCO3(aq) H3C6H5O7(aq) 3CO2(g) 3H2O(l) Na3C6H5O7(aq) How many moles of Na3C6H5O7 can be
produced if one tablet containing 0.0119 mol of NaHCO3 is dissolved?0.0119 mol NaHCO3 (1 mol Na3C6H5O7/3 mol NaHCO3) 0.00397 mol
63. Esterification The process in which an organic acid and an alcohol that forms as ester and water is known as esterification. Ethyl butanoate (C3H7COOC2H5), an ester, is formed when the alcohol ethanol (C2H5OH) and butanoic acid (C3H7COOH) are heated in the presence of sulfuric acid.
C2H5OH(l) C3H7COOH(l) C3H7COOC2H5(l) H2O(l)
Determine the mass of ethyl butanoate produced if 4.50 mol of ethanol is used.
4.50 mol C2H5OH 1 mol C3H7COOC2H5 __
1 mol C2H5OH
116.18 g C3H7COOC2H5 ___
1 mol C3H7COOC2H5
523 g C3H7COOC2H5
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64. Greenhouse Gas Carbon dioxide is a green-house gas that is linked to global warming. It is released into the atmosphere through the combustion of octane (C8H18) in gasoline. Write the balanced chemical equation for the combus-tion of octane and calculate the mass of octane needed to release 5.00 mol of CO2.2C8H18(l) 25O2(g) 0 16CO2(g) 18H2O(l)
5.00 mol CO2 2 mol C8H18 __ 16 mol CO2
0.625 mol C8H18
0.625 mol C8H18 114.28 g C8H18 __
1 mol C8H18 71.4 g C8H18
65. A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate.a. Write the balanced chemical equation.
K2CrO4(aq) Pb(NO3)2(aq) 0 PbCrO4(s) 2KNO3(aq)
b. Starting with 0.250 mol of potassium chromate, determine the mass of lead chromate formed.
0.250 mol K2CrO4 1 mol PbCrO4 __ 1 mol K2CrO4
323.2 g PbCrO4 __ 1 mol PbCrO4
80.8 g PbCrO4
66. Rocket Fuel The exothermic reaction between liquid hydrazine (N2H2) and liquid hydrogen peroxide (H2O2) is used to fuel rockets. The products of this reaction are nitrogen gas and water.a. Write the balanced chemical equation.
N2H2(l) H2O2(l) 0 N2(g) 2H2O(g)
b. How much hydrazine, in grams, is needed to produce 10.0 mol of nitrogen gas?
10.0 mol N2 1 mol N2H2 __
1 mol N2
30.03 g N2H2 __ 1 mol N2H2
3.00 102 g N2H2
67. Chloroform (CHCl3), an important solvent, is produced by a reaction between methane and chlorine.
CH4(g) 3Cl2(g) 0 CHCl3(g) 3HCl(g) How much CH4 in grams is needed to produce
50.0 grams of CHCl3?
50.0 g CHCl3 1 mol CHCl3 __
119.37 g CHCl3
1 mol CH4 __ 1 mol CHCl3
16.04 g CH4 __ 1 mol CH4
6.72 g CH4
68. Oxygen Production The Russian Space Agency uses potassium superoxide (KO2) for the chemical oxygen generators in their space suits.
4KO2 2H2O 4CO2 0 4 KHCO3 3O2 Complete Table 11.3.
Oxygen Generation Reaction Data
Mass KO2
Mass H2O
MassCO2
Mass KHCO3
Mass O2
1100 g 140 g 7.0 102 g 1600 g 380 g
380 g O2 (1 mol O2/32.00 g O2) (4 mol KO2/3 mol O2) (71.1 g KO2/1 mol KO2) 1100 g
380 g O2 (1 mol O2/32.00 g O2) (2 mol H2O/3 mol O2) (18.02 g H2O/1 mol H2O) 140 g
380 g O2 (1 mol O2/32.00 g O2) (4 mol CO2/3 mol O2) (44.01 g CO2/1 mol CO2) 7.0 10
2 g
380 g O2 (1 mol O2/32.00 g O2) (4 mol KHCO3/3 mol O2) (100.12 g KHCO3/1 mol KHCO3) 1600 g
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69. Gasohol is a mixture of ethanol and gasoline. Balance the equation, and determine the mass of CO2 produced from the combustion of 100.0 g of ethanol.
C2H5OH(l) O2(g) 0 CO2(g) H2O(g)C2H5OH(l) 3O2(g) 0 2CO2(g) 3H2O(l)
100.0 g C2H5OH 1 mol C2H5OH __
46.08 g C2H5OH
2.170 mol C2H5OH
2.170 mol C2H5OH 2 mol CO2 __
1 mol C2H5OH
4.340 mol CO2
4.340 mol CO2 44.01 g CO2 __ 1 mol CO2
191.0 g CO2 produced
70. Car Battery Car batteries use lead, lead(IV) oxide, and a sulfuric acid solution to produce an electric current. The products of the reaction are lead(II) sulfate in solution and water.a. Write the balanced chemical equation for
this reaction.Pb(s) PbO2(s) 2H2SO4(aq) 0 2PbSO4(aq) 2H2O(l)
b. Determine the mass of lead(II) sulfate produced when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid.
25 g Pb 1 mol Pb __ 207.2 g Pb
2 mol PbSO4 __
1 mol Pb
303.23 g PbSO4 __
1 mol PbSO4 73.2 g PbSO4
71. To extract gold from its ore, the ore is treated with sodium cyanide solution in the presence of oxygen and water.
4Au(s) 8NaCN(aq) O2(g) 2H2O(l) 0 4NaAu(CN)2(aq) 4NaOH(aq)a. Determine the mass of gold that can be
extracted if 25.0 g of sodium cyanide is used.
25.0 g NaCN 1 mol NaCN __ 49.01 g NaCN
4 mol Au __ 8 mol NaCN
196.97 g Au
__ 1 mol Au
50.2 g Au
b. If the mass of the ore from which the gold was extracted is 150.0 g, what percentage of the ore is gold?
50.2 g Au
__ 150.0 g ore
100 33.5% gold in ore
72. Film Photographic film contains silver bromide in gelatin. Once exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate (Na3Ag(S2O3)2) is produced.
AgBr(s) 2Na2S2O3(aq) 0 Na3Ag(S2O3)2(aq) NaBr(aq)
Determine the mass of Na3Ag(S2O3)2 produced if 0.275 g of AgBr is removed.
0.275 g AgBr 1 mol AgBr
__ 187.77 g AgBr
1.46 103 mol AgBr
1.46 103 mol AgBr 1 mol Na3Ag(S2O3)2 __
1 mol AgBr
1.46 103 mol Na3Ag(S2O3)2
1.46 103 mol Na3Ag(S2O3)2
401.12 g Na3Ag(S2O3)2 ___
1 mol Na3Ag(S2O3)2 0.587 g Na3Ag(S2O3)2
Section 11.3
Mastering Concepts 73. How is a mole ratio used to find the limiting
reactant?The actual mole ratio of reactants from the chemical equation is compared to the mole ratio determined from the given quantities.
74. Explain why the statement the limitingreactant is the reactant with the lowest mass is incorrect.The limiting reactant is the reactant that produces the lowest number of moles of product. Mass does not determine the limiting reactant, but the number of moles.
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75. Figure 11.12 uses squares to represent Element M and circles to represent Element N.a. Write the balanced equation for the reaction.
3M2 N2 0 2M3N
b. If each square represents 1 mol of M and each circle represents 1 mol of N, how many moles of M and N were present at the start of the reaction?6 moles of element M (in the form of 3 moles of M2) and 6 moles of element N (likewise, 3 moles of N2)
c. How many moles of product form? How many moles of M and N are unreacted?2 moles of M3N form with 2 moles of N2 unreacted (4 total moles of element N)
d. Identify the limiting reactant and excess reactant.
M2 is the limiting reactant and N2 is the excess reactant.
Mastering Problems 76. The reaction between ethyne (C2H2) and
hydrogen (H2) is illustrated in Figure 11.13. The product is ethane (C2H6). Which is the limiting reactant? Which is the excess reactant? Explain.
Ethyne Hydrogen Ethane Ethyne
+ +
Hydrogen is limiting; ethyne is the excess reactant. One mol of ethyne is left over.
77. Nickel-Iron Battery In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery.
Fe(s) 2NiO(OH)(s) 2H2O(l) 0 Fe(OH)2(s) 2Ni(OH)2(aq)
How many mol of Fe(OH)2 are produced when 5.00 mol of Fe and 8.00 mol of NiO(OH) react?According to the balanced equation, two moles of NiO(OH) react with each mole of Fe. So, 4 mol Fe react with 8.00 mol NiO(OH), leaving 1.00 mol Fe in excess. For each mole of Fe that reacts, one mole of Fe(OH)2(s) is produced. Because 4.0 mol Fe reacts, 4.0 mol Fe(OH)2 is produced.
78. One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride?
CsF(s) XeF6(s) 0 CsXeF7(s).
10.0 mol XeF6 1 mol CsXeF7 __
1 mol XeF6 10.0 mol CsXeF7
79. Iron Production Iron is obtained commer-cially by the reaction of hematite (Fe2O3) with carbon monoxide. How many grams of iron are produced if 25.0 mol of hematite react with 30.0 mol of carbon monoxide?
Fe2O3(s) 3CO(g) 0 2Fe(s) 3CO2(g)According to the balanced equation, 1 mole of hematite reacts with 3 moles of carbon monoxide. Since 25.0 mol of hematite would require 75.0 mol CO but only 30.0 mol are available, CO is the limiting reactant.
30.0 mol CO 2 mol Fe _ 3 mol CO
55.85 g Fe
_ 1 mol Fe
1120 g Fe
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80. The reaction of chlorine gas with solid phosphorus (P4) produces solid phosphorus pentachloride. When 16.0 g of chlorine reacts with 23.0 g of P4, which reactant is limiting? Which reactant is in excess?10Cl2(g) P4(s) 0 4PCl5(s)
16.0 g Cl2 1 mol CI2 __
70.90 g CI2 0.226 mol CI2
23.0 g P4 1 mol P4 __
123.88 g P4 0.185 moles
According to the balanced equation, Cl2 reacts with P4 in a ten-to-one ratio.
0.226 mol Cl2 1 mol P4 _ 10 mol CI2
0.0226 mol P4 needed
Cl2 is limiting reactant; P4 is in excess.
81. Alkaline Battery An alkaline battery produces electrical energy according to this equation.
Zn(s) 2MnO2(s) H2O(l) 0 Zn(OH)2(s) Mn2O3(s)a. Determine the limiting reactant if 25.0 g of
Zn and 30.0 g of MnO2 are used.
25.0 g Zn 1 mol Zn _ 65.3 g Zn
0.380 mol Zn
30.0 g MnO2 1 mol MnO2 __
86.92 g MnO2
0.345 mol MnO2
According to the balanced equation, MnO2 reacts with Zn in a two-to-one ratio. In the reaction, the ratio is 1:1.1 or 0.345/0.380. MnO2 is the limiting reactant.
b. Determine the mass of Zn(OH)2 produced.
0.345 mol MnO2 1 mol Zn(OH)2 __ 2 mol MnO2
99.39 g Zn(OH)2 __ 1 mol Zn(OH)2
17.1 g Zn(OH)2
82. Lithium reacts spontaneously with bromine to produce lithium bromide. Write the balanced chemical equation for the reaction. If 25.0 g of lithium and 25.0 g of bromine are present at the beginning of the reaction, determine2Li(s) Br2(l) 0 2LiBr(s)
a. the limiting reactant.
25.0 g Li 1 mol Li _ 6.94 g Li
3.60 mol Li
25.0 g Br2 1 mol Br2 __
159.80 g Br2 0.156 mol Br2
Actual ratio of mol Li to mol Br2 is 3.60 mol Li/0.156 Br2 or 23:1. Only 2 mol Li are required for 1 mol Br2. Thus, Br2 is the limiting reactant.
b. the mass of lithium bromide produced.
0.156 mol Br2 2 mol LiBr _ 1 mol Br2
0.312 mol LiBr
0.312 mol LiBr 86.84 g LiBr
__ 1 mol LiBr
27.1 g LiBr
c. the excess reactant and the excess mass.Li is in excess.
0.156 mol Br2 2 mol Li _ 1 mol Br2
0.312 mol Li used
3.60 mol Li 0.312 mol Li used 3.29 mol Li remaining
3.29 mol Li 6.94 g Li
_ 1 mol Li
22.8 g Li remaining
Section 11.4
Mastering Concepts 83. What is the difference between actual yield and
theoretical yield?Actual yield is the amount of product actually obtained experimentally. Theoretical yield is the amount of product predicted by a stoichiometric calculation.
84. How are actual yield and theoretical yield determined?Actual yield is determined through experimentation. Theoretical yield is calculated from a given reactant or the limiting reactant.
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85. Can the percent yield of a chemical reaction be more than 100%? Explain your answer.No, you cannot produce more product than the theoretical yield, which is determined from the starting reactants.
86. What relationship is used to determine the percent yield of a chemical reaction?
actual yeild
__ theoretical yield
100 percent yield
87. What experimental information do you need in order to calculate both the theoretical and the percent yield of any chemical reaction?The quantity of one reactant and the actual yield of the product
88. A metal oxide reacts with water to produce a metal hydroxide. What additional information would you need to determine the percent yield of metal hydroxide from this reaction?the mass of one substance in the reaction and the actual mass of metal hydroxide produced
89. Examine the reaction represented in Figure 11.14. Determine if the reaction went to completion. Explain your answer, and calculate the percent yield of the reaction.
Element AElement B
The reaction did not go to completion. Using squares to represent Element A and circles to represent Element B, the initial products would have yielded four AB2 particles, but only three were produced. There are enough remaining unreacted A and B particles to produce one more AB2 particle. The percent yield is 75%.
Mastering Problems 90. Ethanol (C2H5OH) is produced from the
fermentation of sucrose (C12H22O11) in the presence of enzymes.
C12H22O11(aq) H2O(g) 0 4C2H5OH(l) 4CO2(g)
Determine the theoretical yield and the percent yield of ethanol if 684 g of sucrose undergoes fermentation and 349 g of ethanol is obtained.theoretical yield:
684 g C12H22O11 1 mol C12H22O11 __
342.23 g C12H22O11
4 mol C2H5OH __
1 mol C12H22O11
46.07 g C2H5OH __ 1 mol C2H5OH
369 g C2H5OH
% yield actual yield
__ theoretical yield
100
349 g
_ 369 g
100 94.6%
91. Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is:
PbS(s) O2(g) 0 PbO(s) SO2(g)a. Balance the equation, and determine the
theoretical yield of PbO if 200.0 g of PbS is heated.2PbS 3O2 0 2PbO 2SO2
Theoretical yield 200.0 g PbS 1 mol PbS __ 239.27 g PbS
2 mol PbO __ 2 mol PbS
223.19 g PbO
__ 1 mol PbO
186.6 g PbO
b. What is the percent yield if 170.0 g of PbO is obtained?
% yield 170.0 g
_ 186.6 g
100 91.10%
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92. Upon heating, calcium carbonate (CaCO3) decomposes to produce calcium oxide (CaO) and carbon dioxide (CO2).a. Determine the theoretical yield of CO2 if
235.0 g of CaCO3 is heated.CaCO3(s) 0 CaO(s) CO2(g)
235 g CaCO3 1 mol CaCO3 __
100.06 g CaCO3
1 mol CO2 __ 1 mol CaCO3
43.99 g CO2 __ 1 mol CO2
103 g CO2
b. What is the percent yield of CO2 if 97.5 g of CO2 is collected?
% yield actual yield
__ theoretical yield
100
97.5 g CO2 __ 103 g CO2
100 94.7%
93. Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosi-licic acid (H2SiF6).
SiO2(s) 6HF(aq) 0 H2SiF6(aq) 2H2O(l) 40.0 g SiO2 and 40.0 g HF react to yield 45.8 g
H2SiF6.a. What is the limiting reactant?
40.0 g SiO2 1 mol SiO2 __
60.09 g SiO2 0.666 mol SiO2
40.0 g HF 1 mol HF __ 20.01 g HF
2.00 mol HF
Ratio of HF to SiO2 in the balanced equation is 6:1. Actual ratio of HF to SiO2 is
2.00 mol HF __ 0.666 mol SiO2
or 3.00:1. Thus, HF is the
limiting reactant.
b. What is the mass of the excess reactant?
2.00 mol HF 1 mol SiO2 __ 6 mol HF
0.333 mol SiO2
0.666 mol SiO2 available 0.333 mol SiO2 used 0.333 mol SiO2 in excess
0.333 mol SiO2 60.09 g SiO2 __ 1 mol SiO2
20.0 g SiO2 in excess
c. What is the theoretical yield of H2SiF6?
2.00 mol HF 1 mol H2SiF6 __
6 mol HF
0.333 mol H2SiF6
0.333 mol H2SiF6 144.11 g H2SiF6 __
1 mol H2SiF6
48.0 g H2SiF6
48.0 g H2SiF6 is the theoretical yield.
d. What is the percent yield?
% yield 45.8 g H2SiF6 __ 48.0 H2SiF6
100 95.4% yield
94. Van Arkel Process Pure zirconium is obtained using the two-step Van Arkel process. In the first step, impure zirconium and iodine are heated to produce zirconium iodide (ZrI4). In the second step, ZrI4 is decomposed to produce pure zirconium.
ZrI4(s) 0 Zr(s) 2I2(g) Determine the percent yield of zirconium if
45.0 g of ZrI4 is decomposed and 5.00 g of pure Zr is obtained.
theoretical yield 45.0 g ZrI4 1 mol ZrI4 __
598.84 g ZrI4
1 mol Zr _ 1 mol ZrI4
91.22 g Zr
_ 1 mol Zr
6.85 g of Zr
% yield actual yield
__ theoretical yield
100
5.00 g Zr
_ 6.85 g Zr
100 73.0%
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95. Methanol, wood alcohol, is produced when carbon monoxide reacts with hydrogen gas.
CO 2H2 0 CH3OH
When 8.50 g of carbon monoxide reacts with an excess of hydrogen, 8.52 g of methanol is collected. Complete Table 11.4, and calculate the percent yield for this reaction.
Methanol Reaction Data
CO(g) CH3OH(l)
Mass 8.50 g 9.71 g
Molar mass 28.01 g/mol 32.05 g/mol
Moles 0.303 mol 0.303 mol
8.50 g CO (1 mol CO/28.01 g CO) 0.303 mol CO
0.303 mol CO (1 mol CH3OH/1 mol CO) 0.303 mol CH3OH
0.304 mol CH3OH (32.05 g CH3OH/1 mol CH3OH) 9.71 g CH3OH
Percent yield (8.52 g/9.71 g) 100 87.7% yield
96. Phosphorus (P4) is commercially prepared by heating a mixture of calcium phosphate (CaSiO3), sand (SiO2), and coke (C) in an electric furnace. The process involves two reactions.
2Ca3(PO4)2(s) 6SiO2(s) 0 6CaSiO3(l) P4O10(g)
P4O10(g) 10C(s) 0 P4(g) 10CO(g) The P4O10 produced in the first reaction reacts
with an excess of coke (C) in the second reac-tion. Determine the theoretical yield of P4 if 250.0 g of Ca3(PO4)2 and 400.0 g of SiO2 are heated. If the actual yield of P4 is 45.0 g, deter-mine the percent yield of P4.Step 1 is to determine the excess quantity in equation #1.
400.0 g SiO2 1 mol SiO2 __
60.08 g SiO2 6.657 mol of SiO2
250.0 g Ca3(PO4)2 1 mol Ca3(PO4)2 __
310.17 g Ca3(PO4)2
0.8060 mol
According to the balanced equation, Ca3(PO4)2 reacts with SiO2 in a one-to-three ratio. In this reaction, SiO2 is in excess and 0.8060 mol of Ca3(PO4)2 react.
Step 2 is to determine amount of P4O10 produced.
0.8060 mol Ca3(PO4)2 1 mol P4O10 __
2 mol Ca3(PO4)2
0.4030 mol P4O10
Step 3 is to determine amount of P4 produced in step 2 from 0.4030 mol P4O10.
0.4030 mol P4O10 1 mol P4 __
1 mol P4O10
123.88 g P4 __ 1 mol P4
49.92 g of P4
Step 4 is to determine the percent yield, given actual yield is 49.92 g.
% yield actual yield
__ theoretical yield
100
45.0 g P4 _ 49.92 g P4
100 90.1%
97. Chlorine forms from the reaction of hydro-chloric acid with manganese(IV) oxide. The balanced equation is:
MnO2 4HCl 0 MnCl2 Cl2 2H2O
Calculate the theoretical yield and the percent yield of chlorine if 86.0 g of MnO2 and 50.0 g of HCl react. The actual yield of Cl2 is 20.0 g.Step 1: The balanced equation is given.
MnO2 (s) 4HCl (aq) 0 MnCl2 (aq) Cl2 (g) 2H2O (l)
Step 2 is to determine which reactant is in excess.
86.0 g MnO2 1 mol MnO2 __
86.94 g MnO2 0.989 mol MnO2
50.0 g HCl 1 mol HCI __ 36.46 g HCI
1.37 mol HCI
According to the balanced equation, MnO2 reacts with HCl in a one-to-four ratio. In this reaction, the ratio is 0.989 mol/1.37 mol or 1:1.38. Therefore MnO2 is in excess, and HCl is the limiting reactant.
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Step 3 is to determine the grams of Cl2 produced.
1.37 mol HCl 1 mol CI2 _ 4 mol HCI
70.90 g CI2 __ 1 mol CI2
24.3 g CI2.
Step 4 is to determine percent yield, given that 20.0 g of Cl2 is actually formed.
20.0 g CI2 _ 24.3 g CI2
100 82.3%
Mixed Review 98. Ammonium sulfide reacts with copper(II)
nitrate in a double replacement reaction. What mole ratio would you use to determine the moles of NH4NO3 produced if the moles of CuS are known?(NH4)2S Cu(NO3)2 0 2NH4NO3 CuS
2 mol NH4NO3 __
1 mol CuS
99. Fertilizer The compound calcium cyanamide (CaNCN) can be used as a nitrogen source for crops. To obtain this compound, calcium carbide is reacted with nitrogen at high temperatures.
CaC2(s) N2(g) 0 CaNCN(s) C(s) What mass of CaNCN can be produced if
7.50 mol of CaC2 reacts with 5.00 mol of N2?
5.00 mol N2 1 mol CaNCN __
1 mol N2
80.11 g CaNCN __
1 mol CaNCN
401 g CaNCN
100. When copper(II) oxide is heated in the presence of hydrogen gas, elemental copper and water are produced. What mass of copper can be obtained if 32.0 g of copper(II) oxide is used?CuO H2 0 Cu H2O
32.0 g CuO 1 mol CuO __ 79.55 g CuO
1 mol Cu __ 1 mol CuO
63.55 g Cu
__ 1 mol Cu
25.6 g Cu
101. Nitrogen oxide is present in urban pollution, but it immediately converts to nitrogen dioxide as it reacts with oxygen.
a. Write the balanced chemical equation for the formation of nitrogen dioxide from nitrogen oxide.2NO(g) O2(g) 0 2NO2(g)
b. What mole ratio would you use to convert from moles of nitrogen oxide to moles of nitrogen dioxide?
2 mol NO2 __ 2 mol NO
102. Electrolysis Determine the theoretical and percent yield of hydrogen gas if 36.0 g of water undergoes electrolysis to produce hydrogen and oxygen and 3.80 g of hydrogen is collected.2H2O(l) 0 H2(g) O2(g)
36.0 g H2O (1 mol H2O/18.02 g H2O) 2.00 mol
theoretical yield 2.00 mol H2O 1 mol H2 __ 2 mol H2O
2.02 g H2 _ 1 mol H2
4.04 g H2
% yield 3.80 g H2 _ 4.04 g H2
100 94.1%
103. Iron reacts with oxygen as shown.
500
10 15 25 30 3520
Mas
s of
Fe 2
O3
(g)
30
20
10
Mass of Fe (g)
Mass of Fe2O3 Formed From Burning Fe
4Fe(s) 1 3O2(g) 0 2Fe2O3(s) Different amounts of iron were burned in a
fixed amount of oxygen. For each mass of iron burned, the mass of iron(II) oxide formed was plotted on the graph shown in Figure 11.15. Why does the graph level off after 25.0 g of iron is burned? How many moles of oxygen are present in the fixed amount?
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The graph levels off because the oxygen limits the reaction at that point.
35.0 g Fe2O3 1 mol Fe2O3 __
159.7 g Fe2O3
3 mole O2 __ 2 mole Fe2O3
0.329 mol O2
Think Critically 104. Analyze and Conclude In an experiment,
you obtain a percent yield of product of 108%. Is such a percent yield possible? Explain. Assuming that your calculation is correct, what reasons might explain such a result?No, percent yields cannot be greater than 100%. High results could mean the product was not completely dry, or it was contaminated.
105. Observe and Infer Determine whether each reaction depends on a limiting reactant. Explain why or why not, and identify the limiting reactant.a. Potassium chlorate decomposes to form
potassium chloride and oxygen.No, because there is only one reactant.
b. Silver nitrate and hydrochloric acid react to produce silver chloride and nitric acid.Yes, because there are two reactants. Not enough information is given to identify which is the limiting reactant.
106. Design an Experiment Design an experi-ment that can be used to determine the percent yield of anhydrous copper(II) sulfate when copper(II) sulfate pentahydrate is heated to remove water.Obtain and record the mass of an empty evaporating dish. Add 2.00 g of copper(II) sulfate pentahydrate and obtain and record the mass of the hydrate and evaporating dish. Heat the dish gently for 5 minutes, then strongly for 5 minutes to drive off the water. Cool the dish and remass. Record. Determine the mass of the anhydrous copper sulfate. Using the equation CuSO45H2O 0 CuSO4 5H2O and the initial mass of the copper(II) sulfate pentahydrate, determine the theoretical yield of copper(II)
sulfate. Determine the actual yield of copper(II) sulfate. Divide the actual yield by the theoretical yield and multiply by 100 to determine percent yield of copper(II) sulfate.
107. Apply Concepts When a campfire begins to die down and smolder, the flame can be rekindled if you start to fan it. Explain in terms of stoichiometry why the fire begins to flare up again.When you fan the flame, additional oxygen is added and the remaining coals can burn.
108. Apply Concepts Students conducted a lab to investigate limiting and excess reac-tants. The students added different volumes of sodium phosphate solution (Na3PO4) to a beaker. They then added a constant volume of cobalt(II) nitrate solution (Co(NO3)2), stirred the contents, and allowed the beakers to sit overnight. The next day, each beaker had a purple precipitate formed at the bottom. The students decanted the supernatant from each beaker, divided it into two samples, and added one drop of sodium phosphate solution to one sample and one drop of cobalt(II) nitrate solu-tion to the second sample. Their results are shown in Table 11.5.
Reaction Data for Co(NO3)2 and Na3PO4
TrialVolumeNa3PO4
VolumeCo(NO3)2
Reaction with Drop of Na3PO4
Reaction with
Drop of Co(NO3)2
1 5.0 mL 10.0 mL purple precipitate
no reaction
2 10.0 mL 10.0 mL no reaction
purple precipitate
3 15.0 mL 10.0 mL no reaction
purple precipitate
4 20.0 mL 10.0 mL no reaction
purple precipitate
a. Write a balanced chemical equation for the reaction.2Na3PO4 (aq) 3Co(NO3)2 (aq) 0 Co3(PO4)2 (s) 6NaNO3 (aq)
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b. Based on the results, identify the limiting reactant and the excess reactant for each trial.Trial 1: Na3PO4 is limiting, Co(NO3)2 is in excess because the addition of Na3PO4 caused an additional reaction.
Trials 2-4: Co(NO3)2 is limiting, Na3PO4 is in excess because the addition of Co(NO3)2 caused and additional reaction.
Challenge Problem 109. When 9.59 g of a certain vanadium oxide is
heated in the presence of hydrogen, water and a new oxide of vanadium are formed. This new vanadium oxide has a mass of 8.76 g. When the second vanadium oxide undergoes additional heating in the presence of hydrogen, 5.38 g of vanadium metal forms.a. Determine the empirical formulas for the
two vanadium oxides.
V: 5.38 g __
50.94 g/mol 0.106 mol
O: 4.21 g __
15.999 g/mol 0.263 mol
Divided the smallest molar quantity.
0.106 mol _ 0.106 mol
1
0.263 mol _ 0.106 mol
2.48 2.5
Multiply by the mole ratio by 2.2 (1 mol V: 2.5 mol O) V2O5
V: 5.38 g __
50.94 g/mol 0.106 mol
O: 3.38 g __
15.999 g/mol 0.211 mol
Divided the smallest molar quantity.
0.106 mol _ 0.106 mol
1
0.211 mol _ 0.106 mol
1.99 2
1 mol V: 2 mol O VO2
b. Write balanced equations for the steps of the reaction.V2O5 H2 0 2VO2 H2O
VO2 2H2 0 V 2H2O
c. Determine the mass of hydrogen needed to complete the steps of this reaction.
9.59 g V2O5 __
181.88 g/mol
1 mol H2 __ 1 mol V2O5
2.016 g H2 __ 1 mol H2
0.106 g H2
8.76 g VO2 __ 82.94 g/mol
2 mol H2 _ 1 mol VO2
2.016 g H2 __
1 mol H2 0.426 g H2
Total hydrogen mass 0.106 g 0.426 g 0.532 g H2
Cumulative Review 110. You observe that sugar dissolves more quickly
in hot tea than in iced tea.You state that higher temperatures increase the rate at which sugar dissolves in water. Is this statement a hypoth-esis or a theory? Why? (Chapter 1)a hypothesis, because it is based only on observation, not on data
111. Write the electron configuration for each of the following atoms. (Chapter 5)a. fluorine
[He]2s22p5
b. aluminum[Ne]3s23p1
c. titanium[Ar]4s23d2
d. radon[Xe]6s24f145d106p6
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112. Explain why the gaseous nonmetals exist as diatomic molecules, but other gaseous elements exist as single atoms. (Chapter 8)Diatomic molecules achieve a noble gas electron configuration by forming covalent bonds. The monoatomic gases already have noble gas electron configurations.
113. Write a balanced equation for the reaction of potassium with oxygen. (Chapter 9)4K(s) O2(g) 0 2K2O(s)
114. What is the molecular mass of UF6? What is the molar mass of UF6? (Chapter 10)352.02 amu, 352.02 g/mol
238.03 6(18.998) 352.02
115. Figure 11.16 gives percent composition data for several organic compounds. (Chapter 10)
Percent Composition ofSome Organic Compounds
10
20
30
40
50
Perc
ent
by m
ass
0Ethanol Acetaldehyde Butanoic acidFormaldehyde
Compound name
52.2
13.0
34.840.0
6.7
53.3 54.5
9.1
36.4
54.5
9.1
36.4
%C%H%O
a. How are the molecular and empirical formulas of acetaldehyde and butanoic acid related?They are whole number multiples of one another.
b. What is the empirical formula of butanoic acid?54.5 g C (1 mol/12.01 g) 4.54 mol
9.1 g H (1 mol/1.01 g) 9.0 mol
36.4 g O (1 mol/16.00 g ) 2.28 mol
4.54 mol C/2.28 mol O 2
9.0 mol H/2.28 mol O 4
2.28 mol O/2.28 mol O 1
The empirical formula is C2H4O.
Additional Assessment 116. Air Pollution Research the air pollutants
produced by combustion of gasoline in internal combustion engines. Discuss the common pollutants and the reaction that produces them. Show, through the use of stoichiometry, how each pollutant could be reduced if more people used mass transit.Answers will vary. Common pollutants are SO2, NO, NO2 and O3. Check that stoichiometric calculations account for a decrease in the pollutant.
117. Haber Process The percent yield of ammonia produced when hydrogen and nitrogen are combined under ordinary conditions is extremely small. However, the Haber Process combines the two gases under a set of conditions designed to maximize yield. Research the conditions used in the Haber Process, and find out why the development of the process was of great importance.Answers will vary. Make sure the equation, N2(g) 3H2(g) 2NH3(g) 92 kJ, is included. The aim of the Haber process was to control a reaction so that a large amount of a useful product was yielded quickly. The process was of great importance because the Germans had to come up with a nitrogen compound that could be produced in large amounts.
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Document-Based QuestionsBombardier Beetles Many insects secrete hydrogen peroxide (H2O2) and hydroquinone C6H4(OH)2. Bombardier beetles take this a step further by mixing these chemicals with a catalyst. The result is an exothermic chemical reaction and a spray of hot, irritating chemicals for any would-be predator. Researchers hope to use a similar method to reignite aircraft turbine engines.
Figure 11.17 below shows the chemical reaction that results in the bombardier beetles defensive spray.
Data obtained from: Becker, Bob. April 2006. ChemMatters.24: no.2
H2O2
C6H4O2C6H4(OH)2
H2O O2 Energy
OH
OH
O
O
Hydroquinone Benzoquinone
Catalyst
118. If the bombardier beetle stores 100.0 mg of hydroquinone (C6H4(OH)2) along with 50.0 mg of hydrogen peroxide (H2O2), what is the limiting reactant? Balanced reaction:
2C6H4(OH)2 4H2O2 0 2C6H4O2 6H2O O2
100 mg 50 mg ? mg
C6H4(OH)2 H2O2 0 C6H4O2 H2O O2 energy
100.0 mg C6H4(OH)2 1 103 g __
1 mg
1 mol __ 110.00 g C6H4(OH)2
9.08 104 mol
C6H4(OH)2
50.0 mg H2O2 1 103 g __
1 mg 1 mol __
34.02 g H2O2
1.47 103 mol H2O2
9.08 104 mol C6H4(OH)2 ___
1.47 103 mol H2O2
0.618 mol C6H4(OH)2 __
1 mol H2O2 from reaction vs.
2 mol C6H4(OH)2 __
4 mol H2O2 from eqauation
H2O2 is the limiting reactant.
119. What is the excess reactant, and how many milligrams are in excess?1.47 103 mol H2O2
2 mol C6H4(OH)2 __ 4 mol H2O2
110.12 g C6H4(OH)2 __
1 mol C6H4(OH)2
1 mg __
1 103 g
80.9 mg C6H4(OH)2 used
100.0 mg 80.9 mg 19.1 mg of C6H4(OH)2 in excess
120. How many milligrams of benzoquinone will be produced?
1.47 103 mol H2O2 2 mol C6H4O2 __
4 mol H2O2
108.09 g C6H4O2 __
1 mol C6H4O2
1 mg __
1 103 g
79.4 mg C6H4O2 produced
Standardized Test Practicepages 398399
1. Stoichiometry is based on the law ofa. constant mole ratios.b. Avogadros constant.c. conservation of energy.d. conservation of mass.d
Use the graph below to answer Questions 25.Supply of Various Chemicalsin Dr. Raitanos Laboratory
Na2CO3500.0 gNaCl
700.0 g
Ca(OH)2300.0 g
NaH2PO4350.0 g
KClO3200.0 gAgNO3
100.0 g
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2. Pure silver metal can be made using the reac-tion shown below.
Cu(s) 2AgNO3(aq) 0 2Ag(s) Cu(NO3)2(aq)
How many grams of copper metal will be needed to use up all of the AgNO3 in Dr. Raitanos laboratory?a. 18.70 gb. 37.3 gc. 74.7 gd. 100 ga
100.0 g AgNO3 1 mol AgNO3 __
169.88 g AgNO3
1mol Cu __ 2 mol AgNO3
63.55w g Cu
__ 1 mol Cu
18.70 g Cu
3. The LeBlanc process is the traditional method of manufacturing sodium hydroxide. The equa-tion for this process is as follows.
Na2CO3(aq) Ca(OH)2(aq) 0 2NaOH(aq) CaCO3(s)
Using the amounts of chemicals available in Dr. Raitanos lab, what is the maximum number of moles of NaOH that can be produced?a. 4.05 molb. 4.72 molc. 8.097 mold. 9.43 molc
Find the limiting reactant.
500.0 g Na2CO3 1 mol Na2CO3 __
106.00 g Na2CO3
4.717 mol Na2CO3
300.0 g Ca(OH)2 1 mol Ca2 (OH)2 __ 74.10 g Ca(OH)2
4.049 mol Ca(OH)2
Ca(OH)2 is the limiting reactant.
4.049 mol Ca(OH)2 2 mol NaOH __
1 mol Ca(OH)2
8.097 mol NaOH
4. Pure O2 gas can be generated from the decom-position of potassium chlorate (KClO3):
2KClO3(s) 0 2KCl(s) 3O2(g) If half of the KClO3 in the lab is used and
12.8 g of oxygen gas is produced, what is the percent yield of this reaction?a. 12.8%b. 32.7%c. 65.6%d. 98.0%b
1 _ 2 (200.0 g KCIO3) 100.0 g KCIO3
theoretical yield 100.0 g KCIO3 1 mol KCIO3 __
122.5 g KCIO3
3 mol O2 __
2 mol KCIO3
32.00 g O2 __ 1 mol O2
39.17 g O2
percent actual yield
__ theoretical yield
100 12.8 g
_ 39.17 g
100 32.7%
5. Sodium dihydrogen pyrophosphate (Na2H2P2O7), more commonly known as baking powder, is manufactured by heating NaH2PO4 at high temperatures:
2NaH2PO4(s) 0 Na2H2P2O7(s) H2O(g) If 444.0 g of Na2H2P2O7 is needed, how much
more NaH2PO4 will Dr. Raitano have to buy to make enough Na2H2P2O7?a. 0.00 gb. 94.0 gc. 130.0 gd. 480 gc
444.0 g Na2H2P2O7 1 mol Na2H2P2O7
__ 221.94 g Na2H2P2O7 2 mol NaH2PO4 __
1 mol Na2H2P2O7
119.98 g NaH2PO4 __ 1 mol Na2H2PO4
48.0 g NaH2PO4
480.0 g 350.0 g 130.0 g
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6. Red mercury(II) oxide decomposes at high temperatures to form mercury metal and oxygen gas:
2HgO(s) 0 2Hg(l) O2(g) If 3.55 mol of HgO decomposes to form
1.54 mol of O2 and 618 g of Hg, what is the percent yield of this reaction?a. 13.2%b. 42.5%c. 56.6%d. 86.8%d
theoretical yield O2 3.55 mol HgO 1 mol O2 __
2 mol HgO
1.775 mol O2
percent yield 1.54 mol _ 1.775 mol
100 86.8%
Y
Y YY
YYY
Y
Y
Y ZZZ
Z
ZZZ
Z
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
Z WWW
WW
WWW
WW
WWW
WW
WWW
WW
WWW
WW
WWW
WWY
ZZ
Z
YY
Y
X X X X X X X X X X X X X XX X X X X X X X X X X X X X
1
PERIODIC TABLE
2
3 4 5 6 7 8 9 10 11 12
13 14 15 16 17
18
7. Which elements tend to have the largest atomic radius in their periods?a. Wb. Xc. Yd. Zc
8. Elements labeled W have their valence electrons in which sublevel?a. sb. pc. dd. fb
9. Dimethyl hydrazine (CH3)2N2H2 ignites sponta-neously upon contact with dinitrogen tetroxide (N2O4):
(CH3)2N2H2(l) 2N2O4(l) 0 3N2(g) 4H2O(g) 2CO2(g)
Because this reaction produces an enormous amount of energy from a small amount of reactants, it was used to drive the rockets on the Lunar Excursion Modules (LEMs) of the Apollo space program. If 18.0 mol of dinitrogen tetroxide is consumed in this reaction, how many moles of nitrogen gas will be released?
Set up a mole ratio: 3 mole N2 __
2 moles N2O4
18 moles N2O4 3 moles N2 __
2 moles N2O4
27 moles N2
Use the table below to answer questions 10 and 11.
First Ionization Energy of Period 3 Elements
Element Atomic Number1st Ionization Energy, kJ/mol
Sodium 11 496
Magnesium 12 736
Aluminum 13 578
Silicon 14 787
Phosphorus 15 1012
Selenium 16 1000
Chlorine 17 1251
Argon 18 1521
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236 Chemistry: Matter and Change Chapter 11 Solutions Manual
SOLUTIONS MANUALCHAPTER 11
10. Plot the data from this data table. Place atomic numbers on the x-axis.Data should form an approximately linear relationship with a few jagged edges, similar to Figure 6.16 on page 191.
1210 14 16 18
Firs
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niza
tion
En
ergy
(Kj/m
ol)
200
400
600
800
1000
1200
1400
1600
Atomic Number
Atomic Number vs. First Ionization Energy
11. Summarize the general trend in ionization energy. How does ionization energy relate to the number of valence electrons in an element?Ionization generally increases as you move across a period or row in the periodic table. Elements in the first few families have only 1 or 2 valence electrons, which are relatively easy to remove since this will result in a complete outer shell. Elements on the right side of the periodic table have very high ionization energies because their outer shells are nearly filled, therefore making it more likely for these elements to gain a few electrons rather than lose many.
12. How much cobalt(III) titanate (Co2TiO4), in moles, is in 7.13 g of the compound?a. 2.39 101 molb. 3.10 102 molc. 3.22 101 mold. 4.17 102 molb
7.13 g Co2TiO4 1 mol Co2TiO4 __
229.74 g Co2TiO4 0.0310 mol
3.10 102 mol
Use the pictures below to answer Questions 1317.a.
b.
c.
d.
e.
13. Hydrogen sulfide displays this molecular shape.a
14. Molecules with this shape have four shared pairs of electrons and no lone pairs of electrons.c
15. This shape of molecule is referred to as trigonal planar.b
16. Carbon dioxide displays this molecular shape.d
17. Molecules with this shape undergo sp2 hybridization.b