PERCENT YIELD
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PERCENT YIELD. Frequently the amount of a product made during a chemical reaction is less that that predicted by mass – mass stoichiometry. Percent yield is calculated by dividing the amount actually produced by the theoretical amount as determined by mass – mass stoichiometry and then

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PERCENT YIELD

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Percent yield

PERCENT YIELD

Frequently the amount of a product made during a chemical reaction is

less that that predicted by mass – mass stoichiometry. Percent yield is

calculated by dividing the amount actually produced by the theoretical

amount as determined by mass – mass stoichiometry and then

multiplying by 100.


Percent yield

Percent Yield Sample Problem: What is the percent yield if 17.5

grams of barium hydroxide (Ba(OH)2 when reacting with an excess

of hydrochloric acid actually produces 2.93 grams of water? Note:

put the grams of given substance over the barium hydroxide and

the grams of water actually produced over the water. Since you want

to determine the theoretical amount of water put a “? “over it as well.

?

17.5 g 2.92 g actual

HCl + Ba(OH)2 BaCl2 + H2O

2

2

Step 1: Balance and solve a mass – mass problem for the theoretical

amount of water which can be produced.

17.5 g Ba(OH)2

1 mol Ba(OH)2

2 mol H2O

18 g H2O

____________________________________________________

=

3.68 g H2O

171 g Ba(OH)2

1 mol Ba(OH)2

1 mol H2O

Step 2: Divide the actual by the theoretical and multiply by 100.

2.92 gx 100 =

3.68g

79.6 % yield


Practice problems

Practice Problems

Try to work the next problems on the paper first. Then use the power point to check your work.


Percent yield

Practice Problem 1: What is the percent yield if the decomposition

of 10.9 grams of sodium bicarbonate (NaHCO3) actually yields 1.71

grams of carbon dioxide (CO2)?

?

10.9 g 1.71 g

NaHCO3 Na2CO3 + CO2 + H2O

2

Solve and then check your answer below.

10.9 g NaHCO3

1 mol NaHCO3

1 mol CO2

44.0 g CO2

___________________________________________________

=

2.85 g CO2

84.0 g NaHCO3

2 mol NaHCO3

1 mol CO2

= 60.0 % yield

1.71 gx 100

2.85 g


Percent yield

Practice Problem 2: What is the percent yield if 13.4 g of sulfuric

acid (H2SO4) reacts with an excess of sodium nitrate (NaNO3) to

actually produce 14.8 g of nitric acid (HNO3)?

?

13.4 g

14.8 g actual

H2SO4 + NaNO3 Na2SO4 + HNO3

2

2

Solve and then check your answer below.

_________________________________________________

13.4 g H2SO4

1 mol H2SO4

2 mol HNO3

63.0 g HNO3

=

17.2 g HNO3

98.1 g H2SO4

1 mol H2SO4

1 mol HNO3

86.0 % yield

14.8 gx 100 =

17.2 g


Percent yield

Practice Problem 3: What is the percent yield if 9.16 g of silver nitrate

reacts with an excess of iron (Fe) to actually produce 5.77 grams of

silver?

?

9.16 g 5.77 g actual

Fe + AgNO3 Ag + Fe(NO3)3

3

3

Solve and then check your answer below.

9.16 g AgNO3

____________ ______________________________

1 mol AgNO3

3 mol Ag

108 g Ag

5.82 g Ag

=

170 g AgNO3

3 mol AgNO3

1 mol Ag

5.77 g x 100

5.82 g

= 99.1 % yield


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