section 7.5: percentage yieldcisforchemistry.weebly.com/uploads/9/9/6/8/... · percentage yield=...

Copyright © 2011 Nelson Education Ltd. Chapter 7: Stoichiometry in Chemical Reactions 7.5-1

Section 7.5: Percentage Yield Tutorial 1 Practice, page 338 1. (a) Given: g 23.5Ni =m Required: theoretical yield of nickel carbonyl,

4Ni(CO)m Solution: Step 1. List the given value, the required value, and the corresponding molar masses. Ni(s) + 4 CO(g) → Ni(CO)4(g) 23.5 g

4Ni(CO)m 58.69 g/mol 170.73 g/mol Step 2. Convert mass of given substance to amount of given substance.

nNi= 23.5 g !

1 molNi

58.69 g

nNi= 0.400 41 mol [two extra digits carried]

Step 3. Convert amount of nickel to amount of nickel carbonyl.

nNi(CO)

4

= 0.400 41 molNi

!

1 molNi(CO)

4

1 molNi

nNi(CO)

4

= 0.400 41 mol

Step 4. Convert amount of nickel carbonyl to mass of nickel carbonyl.

mNi(CO)

4

= (0.400 41 mol )170.73 g

1 mol

!

"#$

%&

mNi(CO)

4

= 68.362 g [two extra digits carried]

Statement: The theoretical yield of nickel carbonyl is 68.4 g. (b) actual yield of nickel carbonyl = 61.0 g Use the theoretical yield of nickel carbonyl from (a).

percentage yield =actual yield

theoretical yield!100 %

=61.0 g

68.362 g!100 %

percentage yield = 89.2 %

Therefore, if 61.0 g of nickel carbonyl is collected, the percentage yield is 89.2 %. 2. Given: g 1.61

4CH =m ; g 2.00OH2=m ; actual yield of hydrogen = 5.0 g

Required: theoretical yield, 2H

m ; percentage yield of hydrogen Solution: Step 1. List the given values, the required value, and the corresponding molar masses. CH4(g) + H2O(g) → CO(g) + 3 H2(g) 1.61 g 2.00 g

2Hm

16.05 g/mol 18.02 g/mol 2.02 g/mol


Step 2. Convert mass of given substances to amount of given substances.

nCH

4

= 1.61 g !

1 molCH

4

16.05 g

nCH

4

= 0.100 31 mol [two extra digits carried]

nH

2O= 2.00 g !

1 molH

2O

18.02 g

nH

2O= 0.110 99 mol [two extra digits carried]

Step 3. Determine the limiting reagent.

nH

2O= 0.100 31 mol

CH4

!

1 molH

2O

1 molCH

4

nH

2O= 0.100 31 mol [two extra digits carried]

Since the amount of water present is greater than the amount required to react with the methane, methane is the limiting reagent. Step 4. Convert amount of methane to amount of hydrogen.

nH

2

= 0.100 31 molCH

4

!

3 molH

2

1 molCH

4

nH

2


Step 5. Convert amount of hydrogen to mass of hydrogen.

mH

2

= (0.300 93 mol )2.02 g

1 mol

!

"#$

%&

mH

2

= 0.607 88 g

Step 6. Calculate the percentage yield.



=0.50 g

0.607 88 g!100 %

percentage yield = 82 %

Statement: The theoretical yield of hydrogen is 0.608 g and the percentage yield is 82 %.


Section 7.5 Questions, page 339 1. The theoretical yield is a prediction of the quantity of product expected based on the stoichiometry of the reaction. The actual yield is the quantity of product actually collected when the reaction is complete. 2. The actual yield rarely equals the theoretical yield because a number of factors limit the quantity of product collected. These include the nature of the reaction, experimental conditions, the presence of impurities in the reactants, and competing side reactions. 3. Reducing the number of steps in an experimental procedure reduces the chances of accidental loss of product. The reacting materials may be lost due to measurements, through spillage during the transfer of solutions, by splattering during heating, during the isolation and purification of the product, or due to other handling errors, making the actual yield less than the theoretical yield. 4. (a) Not drying the mixture enough makes the final product heavier than expected, resulting in a higher percentage yield. (b) Adding insufficient sodium hydrogen carbonate has no effect on the percentage yield because the reaction stops when the sodium hydrogen carbonate is consumed by the reaction. The final mass of sodium chloride will be smaller but the percentage yield remains the same. (c) Using impure sodium hydrogen carbonate produces less than the predicted amount of sodium chloride, resulting in a decreased percentage yield. (d) Some of sodium chloride product is lost as the reaction mixture splatters. Since not all sodium chloride produced is collected, the percentage yield is reduced. 5. The presence of dichloromethane suggests that not all of the methane present initially is converted into chloromethane. The percentage yield of chloromethane is therefore reduced. 6. (a) Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq) (b) Given: g 3.00Zn =m Required: theoretical yield of silver metal, Agm Solution: Step 1. List the given value, the required value, and the corresponding molar masses. Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq) 3.00 g Agm 65.41 g/mol 107.87 g/mol Step 2. Convert mass of given substance to amount of given substance.

nZn

= 3.00 g !1 mol

Zn

65.41 g

nZn


Step 3. Convert amount of zinc to amount of silver.

nAg

= 0.045 865 molZn

!

2 molAg

1 molZn

nAg



Step 4. Convert amount of silver to mass of silver.

mAg

= (0.091 730 mol )107.87 g

1 mol

!

"#$

%&

mAg


Statement: The theoretical yield of silver metal is 9.9 g. (c) actual yield of silver metal = 7.2 g Use the theoretical yield of silver metal from (b).



=7.2 g

9.8949 g!100 %


Therefore, if 7.2 g of silver metal is collected, the percentage yield is 73 %. 7. (a) Given: percentage yield of chlorine = 95 %; theoretical yield of chlorine = 142 g Required: actual yield of chlorine Solution:



actual yield = theoretical yield !percentage yield

100 %

= 142 g !95 %

100 %

actual yield = 134.9 g [two extra digits carried]

Statement: The actual yield of chlorine is 135 g. (b) Given: g 134.9

2Cl =m Required: mass of sodium chloride,

NaClm

Solution: Step 1. Balance the chemical equation for the reaction. List the given value, the required value, and the corresponding molar masses. 2 NaCl(aq) + 2 H2O(l) → Cl2(g) + 2 NaOH(aq) + H2(g) NaClm 134.9 g 58.44 g/mol 70.90 g/mol Step 2. Convert mass of given substance to amount of given substance.

nCl

2

= 134.9 g !

1 molCl

2

70.90 g

nCl

2

= 1.903 mol [two extra digits carried]


Step 3. Convert amount of chlorine to amount of sodium chloride.

nNaCl

= 1.903 molCl

2

!2 mol

NaCl

1 molCl

2

nNaCl

= 3.806 mol

Step 4. Convert amount of sodium chloride to mass of sodium chloride.

mNaCl

= (3.806 mol )58.44 g

1 mol

!

"#$

%&

mNaCl

= 220 g

Statement: To produce 135 g chlorine, 220 g of sodium chloride is required. 8. Given: mol 4.00

C=n ; actual yield of methane = 28.0 g

Required: percentage yield of methane Solution: Step 1. List the given value, the required value, and the corresponding molar masses. 2 C(s) + 2 H2O(g) → CH4(g) + CO2(g) 4.00 mol

4CHm

12.01 g/mol 16.05 g/mol Step 2. Convert amount of carbon to amount of methane.

nCH

4

= 4.00 molC!

1 molCH

4

2 molC

nCH

4

= 2.00 mol

Step 3. Convert amount of methane to mass of methane.

mCH

4

= (2.00 mol )16.05 g

1 mol

!

"#$

%&

mCH

4

= 32.1 g

The theoretical yield of methane is 32.1 g. Step 4. Calculate the percentage yield.



=28.0 g

32.1 g!100 %


Statement: The percentage yield of the reaction is 87.2 %.


9. (a) Given: g 3.43NH =m

Required: theoretical yield of ammonium sulfate, 424 SO)(NHm

Solution: Step 1. List the given value, the required value, and the corresponding molar masses. 2 NH3(g) + H2SO4(aq) → (NH4)2SO4(aq) 3.4 g

424 SO)(NHm 17.04 g/mol 132.17 g/mol Step 2. Convert mass of given substance to amount of given substance.

nNH

3

= 3.4 g !

1 molNH

3

17.04 g

nNH

3


Step 3. Convert amount of ammonia to amount of ammonium sulfate.

n(NH

4)

2SO

4

= 0.1995 molNH

3

!

1 mol(NH

4)

2SO

4

2 molNH

3

n(NH

4)

2SO

4


Step 4. Convert amount of ammonium sulfate to mass of ammonium sulfate.

m(NH

4)

2SO

4

= (0.099 75 mol )132.17 g

1 mol

!

"#$

%&

m(NH

4)

2SO

4

= 13.18 g (two extra digits carried)

Statement: The theoretical yield of ammonium sulfate is 13 g. (b) actual yield of ammonium sulfate = 10.4 g Use the theoretical yield of ammonium sulfate from (a).



=10.4 g

13.18 g!100 %


Therefore, if 10.4 g of ammonium sulfate is collected, the percentage yield is 79 %. 10. Given: g 35.0

43OFe =m ; actual yield of iron = 15.0 g Required: percentage yield of iron Solution: Step 1. List the given value, the required value, and the corresponding molar masses. Fe3O4(s) + 2 C(s) → 3 Fe(s) + 2 CO2(g) 35.0 g

Fem

231.55 g/mol 55.85 g/mol


Step 2. Convert mass of given substance to amount of given substance.

nFe

3O

4

= 35.0 g !

1 moln

Fe3O4

231.55 g

nFe

3O

4


Step 3. Convert amount of magnetite to amount of iron.

nFe= 0.151 16 mol

Fe3O

4

!3 mol

Fe

1 molFe

3O

4

nFe= 0.453 48 mol [two extra digits carried]

Step 4. Convert amount of iron to mass of iron.

mFe= (0.5348 mol )

55.85 g

1 mol

!

"#$

%&

mFe= 25.327 g [two extra digits carried]

The theoretical yield of iron is 25.3 g. Step 5. Calculate the percentage yield.



=15.0 g

25.327 g!100 %


Statement: The percentage yield of the reaction is 59.2 %. 11. (a) N2(g) + 3 H2(g) → 2 NH3(g) (b) Given: g 1000g 1.00

2N ==m ; percentage yield of ammonia = 24.5 % Required: actual yield of ammonia Solution: Step 1. List the given value, the required value, and the corresponding molar masses. N2(g) + 3 H2(g) → 2 NH3(g) 1000 g

3NHm

28.02 g/mol 17.04 g/mol Step 2. Convert mass of given substance to amount of given substance.

nN

2

= 1000 g !

1 molN

2

28.02 g

nN

2


Step 3. Convert amount of nitrogen to amount of ammonia.

nNH

3

= 35.689 molN

2

!

2 molNH

3

1 molN

2

nNH

3



Step 4. Convert amount of ammonia to mass of ammonia.

mNH

3

= (71.378 mol )17.04 g

1 mol

!

"#$

%&

mNH

3


This is the theoretical yield of ammonia. Step 5. Calculate the actual yield.




100 %

= 1216.3 g !24.5 %

100 %


Statement: The actual yield of ammonia is 298 g. (c) Given: amount of nitrogen present = 35.689 mol; actual yield of ammonia = 297.99 g Required: amount of nitrogen unreacted Solution: Step 1. List the given value, the required value, and the corresponding molar masses. N2(g) + 3 H2(g) → 2 NH3(g)

2Nn 297.99 g 28.02 g/mol 17.04 g/mol Step 2. Convert mass of given substance to amount of given substance.

nNH

3

= 297.99 g !

1 molNH

3

17.04 g

nNH

3


Step 3. Convert amount of ammonia reacted to amount of nitrogen reacted.

nN

2

= 17.488 molNH

3

!

1 molN

2

2 molNH

3

nN

2


Step 4. Calculate the amount of nitrogen remaining. 35.689 mol – 8.744 mol = 26.945 mol Statement: The amount of nitrogen that remains unreacted is 26.9 mol. (d) Use the amount of unreacted nitrogen in (c).

mN

2

= (26.945 mol )28.02 g

1 mol

!

"#$

%&

mN

2

= 755 g

The mass of nitrogen that remains unreacted is 755 g.


12. Given: percentage yield of ethanoic acid = 85 %; actual yield of ethanoic acid = 2.5 kg Required: mass of carbon monoxide,

COm

Solution: Step 1. Convert 2.5 kg to a mass in grams.

2.5 kg !1000 g

1 kg= 2500 g

Calculate the theoretical yield of ethanoic acid.



theoretical yield =actual yield

percentage yield!100 %

=2500 g

85 %!100 %

theoretical yield = 2941 g [two extra digits carried]

Step 2. List the known value, the required value, and the corresponding molar masses. CH3OH(l) + CO(g) → HC2H3O2(l)

COm 2941 g

28.01 g/mol 60.06 g/mol Step 3. Convert mass of ethanoic acid to amount of ethanoic acid.

nHC

2H

3O

2

= 2941 g !

1 molHC

2H

3O

2

60.06 g

nHC

2H

3O

2


Step 4. Convert amount of ethanoic acid to amount of carbon monoxide.

nCO

= 48.97 molHC

2H

3O

2

!1 mol

CO

1 molHC

2H

3O

2

nCO


Step 5. Convert amount of carbon monoxide to mass of carbon monoxide.

mCO

= (48.97 mol )28.01 g

1 mol

!

"#$

%&

= 1400 g

mCO

= 1.4 kg

Statement: The mass of carbon monoxide required to prepare 2.5 kg of ethanoic acid is 1.4 kg.


13. Given: percentage yield of reaction = 75.2 %; actual yield of hydrogen sulfide = 50.0 g Required: mass of sulfur,

Sm

Solution: Step 1. Calculate the theoretical yield of hydrogen sulfide.



theoretical yield =actual yield

percentage yield!100 %

=50.0 g

75.2 %!100 %

theoretical yield = 66.489 g [two extra digits carried]

Step 2. List the known value, the required value, and the corresponding molar masses. CH4(g) + 4 S(s) → CS2(g) + 2 H2S(g)

Sm 66.489 g

32.07 g/mol 34.09 g/mol Step 3. Convert mass of hydrogen sulfide to amount of hydrogen sulfide.

nH

2S= 66.489 g !

1 molH

2S

34.09 g

nH

2S= 1.9504 mol [two extra digits carried]

Step 4. Convert amount of hydrogen sulfide to amount of sulfur.

nS= 1.9504 mol

H2S!

4 molS

2 molH

2S

nS= 3.9008 mol [two extra digits carried]

Step 5. Convert amount of sulfur to mass of sulfur.

mS= (3.9008 mol )

32.07 g

1 mol

!

"#$

%&

mS= 125 g

Statement: The mass of sulfur required to produce 50.0 g of hydrogen sulfide is 125 g. 14. (a) Given: g 0.25P =m ; percentage yield of reaction = 90.0 % Required: actual yield of tetraphosphorus decoxide Solution: Step 1. List the known value, the required value, and the corresponding molar masses. P4(s) + 5 O2(g) → P4O10(s) 25.0 g

104OPm

123.88 g/mol 283.88 g/mol


Step 2. Convert mass of phosphorus to amount of phosphorus.

nP

4

= 25.0 g !

1 molP

4

123.88 g

nP

4


Step 3. Convert amount of phosphorus to amount of tetraphosphorus decoxide.

nP

4O

10

= 0.201 81 molP

4

!

1 molP

4O

10

1 molP

4

nP

4O

10


Step 4. Convert amount of tetraphosphorus decoxide to mass of tetraphosphorus decoxide.

mP

4O

10

= (0.201 81 mol )283.88 g

1 mol

!

"#$

%&

mP

4O

10


Step 5. Calculate the actual yield of tetraphosphorus decoxide.




100 %

= 57.290 g !90.0 %

100 %


Statement: The mass of tetraphosphorus decoxide collected from the first step is 51.6 g. (b) Given: g 51.561

104OP =m ; percentage yield of reaction = 95.2 % Required: actual yield of phosphoric acid Solution: Step 1. List the known value, the required value, and the corresponding molar masses. P4O10(s) + 6 H2O(l) → 4 H3PO4(aq) 51.561 g

43POHm

283.88 g/mol 98.00 g/mol Step 2. Convert mass of tetraphosphorus decoxide to amount of tetraphosphorus decoxide.

nP

4O

10

= 51.561 g !

1 molP

4O

10

283.88 g

nP

4O

10



Step 3. Convert amount of tetraphosphorus decoxide to amount of phosphoric acid.

nH

3PO

4

= 0.181 63 molP

4O

10

!

4 molH

3PO

4

1 molP

4O

10

nH

3PO

4


Step 4. Convert amount of phosphoric acid to mass of phosphoric acid.

mH

3PO

4

= (0.726 52 mol )98.00 g

1 mol

!

"#$

%&

mH

3PO

4


This is the theoretical yield of phosphoric acid. Step 5. Calculate the actual yield of phosphoric acid.




100 %

= 71.199 g !95.2 %

100 %

actual yield = 67.8 g

Statement: The mass of phosphoric acid produced from the second step is 67.8 g. 15. Answers may vary. Sample answer: Atom economy and percentage yield are both indicators of the efficiency of a chemical reaction. Atom economy refers to the percentage of atoms in the reactants that are found in the products. If a chemical reaction is highly efficient, the atom economy of the reaction should be high. Percentage yield gives the ratio of the actual quantity of product compared to the predicted theoretical quantity.

section 7.5: percentage yieldcisforchemistry.weebly.com/uploads/9/9/6/8/... · percentage yield=...

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