Section 7.5: Percentage Yield - Ms. Palombo's Grade 11 Chemistry

Section 7.5: Percentage Yield

Tutorial 1 Practice, page 338 1. (a) Given: mNi = 23.5 g

Required: theoretical yield of nickel carbonyl, mNi(CO)4

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

m Ni(CO)4

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 1 mol

g

$ %&

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 ! 100 %

theoretical yield

= 61.0 g ! 100 % 68.362 g

percentage yield = 89.2 %

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

2. Given: mCH4 = 1.61 g ; mH2O = 2.00 g ; actual yield of hydrogen = 5.0 g

Required: theoretical yield, mH2 ; 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

mH2

16.05 g/mol 18.02 g/mol

2.02 g/mol

Copyright ? 2011 Nelson Education Ltd.

Chapter 7: Stoichiometry in Chemical Reactions 7.5-1

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

nCH4

= 1.61

g

! 1 molCH4 16.05 g

nCH4 = 0.100 31 mol [two extra digits carried]

nH2O

=

2.00

g

!

1 molH2O 18.02 g

nH2O = 0.110 99 mol [two extra digits carried]

Step 3. Determine the limiting reagent.

nH2O = 0.100 31

molCH4

! 1 molH2O 1 molCH4

nH2O = 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.

nH2

= 0.100 31

molCH4

! 3 molH2 1 molCH4

nH2 = 0.300 93 mol [two extra digits carried]

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

mH2

= (0.300 93

mol

! ) "#

2.02 g 1 mol

$ %&

mH2 = 0.607 88 g Step 6. Calculate the percentage yield. percentage yield = actual yield ! 100 %

theoretical yield

= 0.50 g ! 100 % 0.607 88 g

percentage yield = 82 %

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

Copyright ? 2011 Nelson Education Ltd.

Chapter 7: Stoichiometry in Chemical Reactions 7.5-2

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: mZn = 3.00 g Required: theoretical yield of silver metal, mAg

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

m Ag

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 molZn 65.41 g

nZn = 0.458 65 mol [two extra digits carried]

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

nAg

= 0.045 865

molZn

! 2 molAg 1 molZn

nAg = 0.091 730 mol [two extra digits carried]

Copyright ? 2011 Nelson Education Ltd.

Chapter 7: Stoichiometry in Chemical Reactions 7.5-3

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

mAg

= (0.091 730

mol

)

! "#

107.87 g 1 mol

$ %&

mAg = 9.8949 g [two extra digits carried]

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). percentage yield = actual yield ! 100 %

theoretical yield

= 7.2 g ! 100 % 9.8949 g

percentage yield = 73 %

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: percentage yield = actual yield ! 100 %

theoretical yield

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: mCl2 = 134.9 g

Required: mass of sodium chloride, mNaCl

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)

m NaCl

134.9 g

58.44 g/mol

70.90 g/mol

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

nCl2

= 134.9

g

! 1 molCl2 70.90 g

nCl2 = 1.903 mol [two extra digits carried]

Copyright ? 2011 Nelson Education Ltd.

Chapter 7: Stoichiometry in Chemical Reactions 7.5-4

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

nNaCl = 1.903

molCl2

! 2 molNaCl 1 molCl2

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: nC = 4.00 mol ; 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

mCH4

12.01 g/mol

16.05 g/mol

Step 2. Convert amount of carbon to amount of methane.

nCH4

= 4.00

molC

! 1 molCH4 2 molC

nCH4 = 2.00 mol

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

mCH4

= (2.00

mol

! ) "#

16.05 g 1 mol

$ %&

mCH4 = 32.1 g The theoretical yield of methane is 32.1 g. Step 4. Calculate the percentage yield. percentage yield = actual yield ! 100 %

theoretical yield

= 28.0 g ! 100 % 32.1 g

percentage yield = 87.2 %

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

Copyright ? 2011 Nelson Education Ltd.

Chapter 7: Stoichiometry in Chemical Reactions 7.5-5

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