1 dragonfly 4 wings - sacramento · pdf filep v t p v = stp ... calculate the theoretical...
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SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P1
Quantity relationships: How much? How many? How many dragonflies’ wings are required for 4 dragonflies to be able to fly?
1 Dragonfly → 4 wings
0
0?
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P2
?4
51 wings
DFsDF
= Same UnitsSame Units
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P3
Quantity Relationships in Chemical Reactions: How to solve stoichiometry problems: What we need to know are two subjects as:
1) A balanced chemical reaction. 2) Relationship between mass and mole.
EXAMPLE: How many moles of oxygen are required to burn 2.40 moles of ethane gas?
2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (l) Given Wanted
0
2 mol C2H6
7 mol O2
? 0
2.40 molC2H6
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P4
?7
4.22 mol
molmol
=
2 mol (C2H6) × ? = 7 mol (O2) × 2.4 mol (C2H6)
molmolmol 4.2
27? ×=
*********************************************** EXAMPLE: Calculate the number of grams of oxygen that are required to burn 155 g of ethane.
2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (l) Given Wanted
? 0
155 g C2H6
)(06.30
155
molg
g
Same Units Same Units
7 mol O2 0
2 molC2H6
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P5
?7
15.52 mol
molmol
=
2715.5? ×=
? = 18.03 mol (oxygen)
Same Units Same Units
Wanted in gram!
massmolarmassmole =
molar mass of oxygen = 2 × 16.00 (g/mol)
? = 18.03 mol × 32.00 (g/mol) ? = 577 g O2
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P6
EXAMPLE: How many milligrams of Nickel Chloride are in a solution if 503 mg of Silver Chloride is precipitated in the reaction of Silver Nitrate and Nickel Chloride solution? 2 AgNO3 (aq) + NiCl2 (aq) → 2 AgCl (s) + Ni(NO3)2 (aq)
Wanted Given
0
1 mol NiCl2
2 molAgCl
(mg) to (g): 503 mg × (1 g/ 1000 mg) = 0.503 g
? 503 mg mg (AgCl) 0 AgCl
mol (AgCl)
?Same Units
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P7
(mass) to (mole) mole = mass / molar mass mole of AgCl = 0.503 g ÷ 143.35 (g/mol) mole of AgCl = 3.51 × 10-3 mol
molmolmol1.72
2?
1=
AgClmolNiClmolAgClmol
211051.3? 23 ××= −
? = 1.75 × 10-3 mol NiCl2
Wanted in milligram!
massmolarmassmole =
molar mass of NiCl2= 129.59 (g/mol)
Mass of NiCl2 = 1.75 × 10-3
mol × 129.59 (g/mol) =0.227 g Mass of NiCl2 = 227 mg
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P8
Gas Stoichiometry at Standard Temperature and Pressure:
One mole of any gas at STP occupies 22.4 LEXAMPLE: What volume of hydrogen, at STP, can be released by 42.7 g of zinc as it reacts with hydrochloric acid?
Zn(s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq)
Given Wanted
?4.22
653.01 L
molmol
=
0
1 mol 22.4 L Zn
? 042.7 g
)(39.65
7.42
molgg
Same Units
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P9
EXAMPLE: 1.30L of gaseous ethylene is burned. What volume of oxygen is required if both gas volume are measured at STP?
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2 H2O(l) Given Wanted
0
1 mol C2H4
3 mol O2
022.4 L 3 × 22.4 L
?2.67
30.14.22
=LL
? 01.30 L
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P10
Gas Stoichiometry at nonstandard condition:
In STP condition: T= 273 K, and P =760 torr
2
22
1
11
TVP
TVP
=
STP = nonstandard
In the previous example for the reaction: Zn (s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq)
We calculated the volume of hydrogen as 14.6 L, now let’s change the problem. Instead of measuring the hydrogen volume at STP, let’s use t = 21 °C and P = 748 torr.
TK=21+273 =294
294748
2736.14760 2
1
V=
×
STP = nonstandard
V2= 16.0 L
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P11
14 transistors
22 resistors 18 capacitors
How many of the above electronic circuits (ec) can you assemble from 22 resistors, 18 capacitors, and 14 transistors?
Calculate the theoretical yield of ec:
ececTecR 55.5
4122 ≈=×
ecCecC 6
3118 =×
Resistor is the limiting item, because 22 of resistors give 5 ec.
→2 + 4 + 3
ecTecT 7
2114 =×
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P12
Limiting Reactants: Example: The fertilizer ammonium nitrate can be made by direct combination of ammonia with nitric acid. If 74.4 g of ammonia is reacted with 159 g of nitric acid, how many grams of ammonium nitrate can be produced? Also calculate the mass of unreacted reactant that is in excess.
NH3 + HNO3 → NH4NO3
Convert the number of grams of each reactant to moles.
NH3: molmol
gg 37.4
03.17
4.74= HNO3: mol
molgg 52.2
02.63
159=
343
343 37.4
11
37.4 NONHmolNHmol
NONHmolNHmol =×
343
343 52.2
1152.2 NONHmol
HNOmolNONHmolHNOmol =×
∴ The reactant that yields the smaller amount of product is limiting reactant.
1 mol of NH3 ≡ 1 mol of HNO3
2.52 mol ≡ ? ? =2.52 mol of HNO3
4.37 mol – 2.52 mol = 1.85 mol NH3 left
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P13
1.85 × 17.03 = 31.51 g NH3 left
A mixture of 5.0 g of H2 (g) and 10.0 g of O2(g) is
ignited. Water forms according to the following
combination reaction:
2H2(g) +O2(g) → 2H2O(g)
Which reactant is limiting? How much water will the
reaction produce?
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P14
Percent Yield: The calculated amount of product if is based on the assumption that all of the reactant is converted into product is called the theoretical yield. In laboratory or in industrial production, the actual amount of product isolated from a reaction is usually less than the theoretical yield, and it is called actual yield. General solution:
αA + βB → γC Given: n gram of A Given: actual yield Wanted: %yield Step 1: Convert mass of A into the mole of A. Step 2: Calculate the theoretical yield:
AmolCmolAmolesofNumber
αγ
×
Step 3: Convert mole of C into the mass of C.
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P15
Step 4:
100% ×=yieldltheoretica
yieldactualYield
Example:
A general chemistry student, preparing copper metal by the reaction of 1.274 g of copper sulfate with zinc metal, obtained a yield of 0.392 g copper. What was the percent yield?
To calculate %yield, we need both the theoretical yield of copper and the actual yield.
CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq)
Mass to Mole:
434 10982.7
6.159
274.1 CuSOmol
molgCuSOg −×=
The theoretical yield:
CumolCuSOmol
CumolCuSOmol 3
44
3 10982.71
110982.7 −− ×=××
Mole to Mass: Cugmol
gCumol 5072.055.6310982.7 3 =×× −
%3.771005072.0392.0100% =×=×=
gg
yieldltheoreticayieldactualYield
SCC-CH110/UCD-CH41 Instructor: J.T. P16 C Chapter: 9
Liquid Oxygen
Liquid Hydrogen
Main Engines
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P17
Thermochemical Equations:
H2 (l) + O2 (l) → H2O (g) + energy
Gives off heat ExothermicSystem -ΔH
2 NH3 (g) + energy → N2(g) + 3 H2 (g)
System Absorbs heat Endothermic +ΔH
The SI unit for energy: joule (J) How many kilojoule of energy are released when 73.0 g C2H6 burns according to: 2C2H6 (g) +7 O2 (g) → 4 CO2(g) + 6H2O (l)+ 3119KJ
0
2 mol C2H6
3119 KJ
?
073.0 g
)(05.30
0.
molgg
73
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P18
ΔH =2.82 × 103 KJ for the photosynthesis reaction by which plants use energy from the sun to form one mole sugar from carbon dioxide and water. How much energy is required to form 454g (1lb) of simple sugar C6H12 O6?
6 CO2 (g) +6 H2O (g) + → C6H12O6 (s) + 6O2 (g)
01 molSuga
2.82 × 103
KJ
?0454 g
)(180
454
molgg
∴ 7.11 × 103 KJ
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P19
SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P20