3.3.5 Theoretical Yield and Percent Yield

Question 1

Theoretical Yield and Percent Yield

Answer 1

• Theoretical yield is the maximum amount of product that can be made from a given amount of reactants.
• Percent yield is the actual yield expressed as a percentage of the theoretical yield

Question 2

note

Answer 2

• Theoretical yield is the maximum amount of product that can be made from a given amount of reactants.
• To calculate theoretical yield, first determine the limiting reactant. Convert the masses of the reactants into moles. Use the amount of each reactant and the coefficients from the balanced chemical equation to determine the limiting reactant. In this example, because the reactants are in a 1:1 ratio, the limiting reactant is ethanol (0.435 moles).
• Finally, calculate the theoretical yield of ethyl acetate by determining the moles produced (using the molar ratio with ethanol, a simple 1:1 in this case). Convert the moles of ethyl acetate to grams using the molar mass, as shown, to obtain the theoretical yield of ethyl acetate in grams (38.3 g).
• Often, in the lab, chemical reactions do not produce their maximum yield. The amount of product produced in an experiment is the actual yield.
• Percent yield is the actual yield expressed as a percentage of the theoretical yield. Divide the actual yield by the theoretical yield and multiply by 100 to obtain the percent yield.
• The value of the actual yield is always less than the value of the theoretical yield and thus the percent yield is always less than 100 %.
• Good chemists generally get higher percent yields

Question 3

How many grams of Na2SeO3 can be made from 5.2 g of SeO2 and 10.0 g sodium hydroxide?

SeO2 + 2NaOH → Na2SeO3 + H2O

Answer 3

8.1 g

Question 4

If the actual yield of H3PO3 in the reaction below is 7.40 g and the theoretical yield is 7.87 g, what is the percent yield?

P4O6 + 6H2O -> 4H3PO3

Answer 4

94.0%

Question 5

Balance the equation and calculate the mass of O2 produced during the thermal decomposition of 100 g of potassium chlorate, KClO3.

KClO3 → KCl + O2

Answer 5

39.2 g O2

Question 6

Balance the following equation, then calculate the theoretical yield of O2 (in moles) that is produced from 3.4 moles of KClO3?

KClO3(s) → KCl(s) + O2(g)

Answer 6

5.1 mol O2

Question 7

What is the percent yield of water produced when 138 g ethyl alcohol, C2H5OH, is combusted in excess oxygen, if the actual experiment generated 150 g water?

C2H5OH + 3O2 -. 2CO2 + 3H2O

Answer 7

92.6%

Question 8

Calcium hydride, CaH2, reacts with water to form calcium hydroxide, Ca(OH)2, and hydrogen gas. How much calcium hydride is needed to form 8.0 g of H2?
CaH2 + 2H2O → Ca(OH)2 + 2H2

Answer 8

84 g CaH2

Question 9

If you obtained 2.50 moles of CuSO4 from the reaction of CuO and H2SO4, but the theoretical yield was 3.19 moles, what was the percent yield?

Answer 9

78.4%

Question 10

What mass of ozone could be produced from a car’s exhaust if 4.00 g NO2 react in an excess of oxygen?

NO2(g) + O2(g) → NO(g) + O3(g)

Answer 10

4.17 g O3

Question 11

If you react 2.0 g AgNO3 and 2.5 g Na2SO4, what is the percent yield if the actual yield is 1.6 g Ag2SO4?

2AgNO3 + Na2SO4 → Ag2SO4 + 2NaNO3

Answer 11

87%

Question 12

What is the theoretical yield (in moles) of MgO that can be formed if you react 3.2 mol of O2 with an excess of magnesium?

2Mg(s) + O2(g) → 2MgO(s)

Answer 12

6.4 mol MgO