## General Chemistry

In the previous post, we talked about the limiting reactant and its importance in determining how much product can be produced in a chemical reaction. The amount of product(s) we calculate based on the limiting reactant is called the theoretical yield.

For example, we saw that 1.6 mol of NO can be obtained when 2 mol of NH3 and 2 mol of O2 are reacted according to the following equation:

4NH3 + 5O2 → 4NO + 6H2O

For this, we calculate separately, how many moles of it can 2 mol of NH3 and 2 mol of O2 produce:

So, this is the theoretical yield that we’d obtain if everything in the reaction worked ideally as written on the paper.

However, most of the time, we do not obtain as much product as the theoretical yield suggests. The amount of product we obtain is the actual yield of the reaction. The theoretical yield and the actual yield of a reaction are often different.

To quantify this difference, the percent yield is used which is the ratio of the actual and theoretical yield multiplied by 100%:

For example: In a chemistry experiment, a student obtained 5.68 g of a product. What is the percent yield of the product if the theoretical yield was 7.12 g?

5.68 g is the actual yield because that is how much material the student obtained. The theoretical yield is 7.12 g which had been calculated based on the limiting reactant.

The percent yield is the ratio of these two:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{10}}{\rm{.68}}\;{\rm{g}}}}{{{\rm{17}}{\rm{.8}}\;{\rm{g}}}}\; \times \;100\% \; = \;60.0\;\%$

An analogy of the percent yield can be comparing how many people make it to the finish-line of a marathon compared to how many had started it. For example, if 1000 people started the marathon and only 600 made to the end, the yield or the efficiency of the marathon is 60%:

# Solving Percent Yield Problems

At this point of studying stoichiometry, we have already learned about molar ratios, stoichiometric calculations based on these ratios, limiting reactant, and the percent yield.

Follow these steps to determine the percent yield or, in general, working on stoichiometry problems:

1) Determine the moles of reactants

2) Determine the limiting reagent

3) Determine the theoretical yield

4) Determine the reaction/percent yield

For example: What is the yield of the reaction if 10.68 g sulfur was obtained by reacting 12.6 g H2S with 14.6 g SO2 according to the following equation?

2H2S(g) + SO2(g) → 3S(s) + 2H2O(g)

1) Determine the moles of reactants:

${\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S)}}\;{\rm{ = }}\;{\rm{12}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{34}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.370}}\;{\rm{mol}}$

${\rm{n}}\;{\rm{(S}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{14}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{64}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.228}}\;{\rm{mol}}$

2) Determine the limiting reagent:

${\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.370 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}$

${\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.228 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.684}}\;{\rm{mol}}$

0.370 mol H2S gives less sulfur and therefore, it is the limiting reactant.

3) Determine the theoretical yield:

H2S gives less S and therefore, it is the LR and 0.555 mol of S is the theoretical yield. The mass of sulfur is:

${\rm{m}}\;{\rm{(S)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{17}}{\rm{.8}}\;{\rm{g}}$

4) Determine the reaction/percent yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{10}}{\rm{.68}}\;{\rm{g}}}}{{{\rm{17}}{\rm{.8}}\;{\rm{g}}}}\; \times \;100\% \; = \;60.0\;\%$

Check Also

#### Practice

1.

What is the theoretical yield of Ca(OH)2, in grams, if 39.27 g of CaO is hydrolyzed (reacted) in an excess of water?

51.9 g Ca(OH)2

Solution

2.

In a chemistry experiment, a student obtained 5.68 g of a product. What is the percent yield of the product if the theoretical yield was 7.12 g?

79.8 %

Solution

The percent yield of the reaction is the ratio of the actual over the theoretical yield. What we obtain is the actual yield and the maximum amount of product that can be obtained from the given amount of the limiting reactant is the theoretical yield.

So, in this example, the actual yield is 5.68 g, and the theoretical yield is 7.12 g.

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{5}}{\rm{.68}}\;{\rm{g}}}}{{{\rm{7}}{\rm{.12}}\;{\rm{g}}}}\; \times \;100\% \; = \;79.8\;\%$

3.

When 38.45 g CCl4 is reacted with an excess of HF, 21.3 g CCl2F2 is obtained. Calculate the theoretical and percent yields of this reaction.

CCl4 + 2HF → CCl2F2 + 2HCl

70.5%

Solution

To find the theoretical yield, we need to calculate the moles of CCl4 and, using the mole ratio, determine how much CCl2F2 can be formed.

${\rm{n}}\;{\rm{(CC}}{{\rm{l}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{38}}{\rm{.45}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{153}}{\rm{.8}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;{\rm{mol}}$

The amount of CCl2F2 that can be formed from this is also 0.250 mol because of the 1:1 molar ratio:

${\rm{n}}\;\left( {{\rm{CC}}{{\rm{l}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.25 }}\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.25}}\;{\rm{mol}}$

The mass, which is also the theoretical yield of CCl2F2 is:

${\rm{m}}\;{\rm{(CC}}{{\rm{l}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{120}}{\rm{.9}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{30}}{\rm{.2}}\;{\rm{g}}$

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{21}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{30}}{\rm{.2}}\;{\rm{g}}}}\; \times \;100\% \; = \;70.5\;\%$

4.

Iron(III) oxide reacts with carbon monoxide according to the equation:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

What is the percent yield of this reaction if 623 g of iron oxide produces 341 g of iron?

78.4%

Solution

The actual yield of the reaction is given as 341 g Fe so, to determine the percent yield, we need to first calculate the theoretical yield. This is the amount of Fe that can be formed if all the starting material (in this case Fe2O3) converts into a product according to the chemical equation.

So, we are going to calculate the moles of 623 g Fe2O3 and determine the moles and the mass of the iron based on the stoichiometric ratio:

${\rm{n}}\;{\rm{(F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{623}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{159}}{\rm{.7}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.90}}\;{\rm{mol}}$

The amount of Fe that can be formed from this is calculated using the mole ratio of the oxide and iron:

${\rm{n}}\;\left( {{\rm{Fe}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.90 }}\cancel{{{\rm{mol}}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Fe}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.80}}\;{\rm{mol}}$

The mass, which is also the theoretical yield of Fe is:

${\rm{m}}\;{\rm{(Fe)}}\;{\rm{ = }}\;{\rm{7}}{\rm{.80}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{435}}\;{\rm{g}}$

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{341}}\;{\rm{g}}}}{{{\rm{435}}\;{\rm{g}}}}\; \times \;100\% \; = \;78.4\;\%$

5.

Determine the percent yield of the reaction if 77.0 g of CO2 are formed from burning 2.00 moles of C5H12 in 4.00 moles of O2.

C5H12 + 8 O2 → 5CO2 + 6H2O

70.0%

Solution

When the amount of both reactants is given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed. This would be the theoretical yield of CO2.

Now, we are given the moles of the reactants, so we calculate how much CO2 can each produce:

${\rm{n}}\;\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}}}\;{\rm{ = }}\;{\rm{10}}{\rm{.0}}\;{\rm{mol}}$

${\rm{n}}\;\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{4}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{8}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{mol}}$

Because oxygen gives less product, it is the LR, and therefore, 2.50 mol CO2 could be formed in this reaction. This is the theoretical yield which is grams is:

${\rm{m}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{44}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{110}}{\rm{.}}\;{\rm{g}}$

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{77}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{110}}{\rm{.}}\;{\rm{g}}}}\; \times \;100\% \; = \;70.0\;\%$

6.

The percent yield for the following reaction was determined to be 84%:

N2(g) + 2H2(g) → N2H4(l)

How many grams of hydrazine (N2H4) can be produced when 38.36 g of nitrogen reacts with 6.68 g of hydrogen?

36.8 g

Solution

The problem asks us to find the actual yield of the reaction given that the percent yield is 80%.

When the amounts of both reactants are given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed. This would be the theoretical yield of hydrazine, and we can use it with the percent yield to find the actual yield of the reaction.

${\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{38}}{\rm{.36}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{28}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}$

${\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{6}}{\rm{.68}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.00}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.34}}\;{\rm{mol}}$

Now, we are given the moles of the reactants, so we calculate how much N2H4 can each produce:

${\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.37 }}\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}$

${\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.34 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.67}}\;{\rm{mol}}$

N2 gives less hydrazine so, it is the LR, and therefore the theoretical yield of N2H4 is 1.37 mol. The mass of N2H4 is:

${\rm{m}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}_{\rm{2}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{43}}{\rm{.8}}\;{\rm{g}}$

The percent yield of the reaction is the ratio of the actual over the theoretical yield, so rearranging this equation we can calculate the actual yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;$

A = T x % Yield

A = 43.8 x 0.84 = 36.8 g

Another approach for determining the actual yield is to convert the moles of limiting reactant to the percentage of the percent yield. So, if the yield is 84%, we could have multiplied the moles of N2 by 0.84 and determine the mass of N2H4 based on that.

We would have 1.37 mole x 0.84 = 1.1508 mol N2 which is the moles of nitrogen that actually convert into hydrazine. By the mole ratio, the moles of N2H4 would be:

${\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.1508 }}\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.1508}}\;{\rm{mol}}$

And the mass would be:

${\rm{m}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}_{\rm{2}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.1508}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.8}}\;{\rm{g}}$

7.

Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams of AgNO?

85%

Solution

The actual yield of the reaction is given as 71.5 g Ag so, to determine the percent yield, we need to first calculate the theoretical yield. This is the amount of Ag that can be formed if all the starting material (in this case AgNO3) converts into product according to the chemical equation.

So, we are going to calculate the moles of 132.5 g AgNO3 and determine the moles and the mass of the silver based on their stoichiometric ratio:

${\rm{n}}\;{\rm{(AgN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{132}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{169}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}$

${\rm{n}}\;\left( {{\rm{Ag}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.780 }}\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Ag}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}$

The mass, which is also the theoretical yield of Ag is:

${\rm{m}}\;{\rm{(Ag)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{107}}{\rm{.9}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{84}}{\rm{.2}}\;{\rm{g}}$

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{71}}{\rm{.5}}\;{\rm{g}}}}{{{\rm{84}}{\rm{.2}}\;{\rm{g}}}}\; \times \;100\% \; = \;85\;\%$

8.

Industrially, nitric acid is produced from ammonia by the Ostwald process in a series of reactions:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

2NO(g) + O2(g) → 2NO2(g)

2NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)

Considering that each reaction has an 85% percent yield, how many grams of NH3 must be used to produce 25.0 kg of HNO3 by the above procedure?

2.20 x 104 g

Solution

Because we are given the mass of the product and need to determine the mass of the starting material, the calculations are going to be in reverse order. So, first, we can calculate the moles of HNO3 and using the mole ratio, find the moles of NO2 which is the link between the 3rd and 2nd reactions.

Before using the mass of HNO3, convert it to grams because the molar mass is given in grams per mole:

${\rm{m}}\left( {{\rm{HN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{25}}{\rm{.0}}\;\cancel{{{\rm{kg}}}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.5}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}$

${\rm{n}}\;{\rm{(HN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{ \times }}\;{10^4}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{63}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{397}}\;{\rm{mol}}$

${\rm{n}}\;\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{397 }}\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{794}}\;{\rm{mol}}$

Now, this would have been the conversion if it was a process with a 100% yield. However, because each step of this process has an 85% percent yield, we need to multiply each mole conversion by a factor of

$\frac{{{\rm{100\% }}}}{{{\rm{85\% }}}}\;{\rm{or}}\;{\rm{simply}}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}$

So, the moles of NO2 that produced 397 mol of HNO3 would be:

${\rm{n}}\;\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{794}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{934}}\;{\rm{mol}}$

Now, we go to the second reaction and see how much NO was needed to produce 934 mol NO2 considering that the yield of the reaction is again 85%.

2NO(g) + O2(g) → 2NO2(g)

${\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{934 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}}}\; \times \;\frac{{100}}{{85}}\;{\rm{ = }}\;{\rm{1098}}\;{\rm{mol}}$

Next, we do the same for the first reaction and determine how much NH3 was needed to produce 1098 mol NO considering that the yield of the reaction is again 85%.

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

${\rm{n}}\;\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{1098 }}\cancel{{{\rm{mol}}\;{\rm{NO}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{NO}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{1293}}\;{\rm{mol}}$

In the last step, we convert the moles to the mass of ammonia:

${\rm{m}}\;{\rm{(N}}{{\rm{H}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{1293}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{17}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{21,981}}\;{\rm{g}}$

Rounding off to three significant figures, we get:

2.20 x 104 g

9.

Aspirin (acetyl salicylic acid) is widely used to treat pain, fever, and inflammation. It is produced from the reaction of salicylic acid with acetic anhydride. The chemical equation for aspirin synthesis is shown below:

In one container, 10.00 kg of salicylic acid is mixed with 10.00 kg of acetic anhydride.

a) Which reactant is limiting? Which is in excess?
b) What mass of excess reactant is left over?
c) What mass of aspirin is formed assuming 100% yield (Theoretical yield)?
d) What mass of aspirin is formed if the reaction yield is 70.0% ?
e) If the actual yield of aspirin is 11.2 kg, what is the percent yield?
f) How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?

a)

Salicylic acid is the limiting reagent.

b)

2.608 x 103 g Acetic anhydride

c)

1.305 x 104 g Aspirin

d)

9.14 x 103 g Aspirin

e)

85.8 %

f)

18.1 kg

Solution

a) Which reactant is limiting? Which is in excess?

The limiting reactant is the one producing less product, so we need to calculate the moles of 00 kg salicylic acid and 10.00 kg acetic anhydride and see which one produces less aspirin. Because the mole ratio of all the reactants and products is 1:1, whichever has a smaller number of moles, it is the limiting reactant.

Before using the masses, convert them to grams because the molar mass is given in grams per mole:

${\rm{m}}\left( {{\rm{salicylic acid}}} \right)\;{\rm{ = }}\;{\rm{10}}{\rm{.00}}\;\cancel{{{\rm{kg}}}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}$

${\rm{m}}\left( {{\rm{acetic anhydride}}} \right)\;{\rm{ = }}\;{\rm{10}}{\rm{.00}}\;\cancel{{{\rm{kg}}}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}$

${\rm{n}}\;{\rm{(salicylic acid)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{138}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}$

${\rm{n}}\;{\rm{(acetic anhydride)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{102}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{97}}{\rm{.9}}\;{\rm{mol}}$

Salicylic acid is present in a smaller quantity, and because of the 1:1 molar ratio of all the reactants and products, it is going to produce less aspirin, and therefore, it is the limiting reactant.

b) What mass of excess reactant is left over?

We know that salicylic acid is the limiting reactant, so the calculations are going to be based on its moles (72.4 mol). Using the mole ratio, we can calculate how much acetic anhydride has reacted. The remaining mass of the acetic anhydride is calculated by subtracting this amount from the initial mass.

Again, because of the 1:1 molar ratio, there is going to be 72.4 mol acetic anhydride reacted:

${\rm{n}}\;\left( {{\rm{acetic anhydride}}} \right)\;{\rm{ = }}\;{\rm{72}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{acetic anhydride}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}$

This corresponds to:

${\rm{m}}\;{\rm{(acetic anhydride)}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{102}}{\rm{.1}}\;{\rm{g}}\;}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{7392}}\;{\rm{g}}$

The remaining mass of acetic anhydride is:

10,000 – 7392 = 2608 g = 2.608 kg

c) What mass of aspirin is formed assuming 100% yield (Theoretical yield)?

72.4 mol aspirin will be formed because of the 1:1 molar ratio:

${\rm{n}}\;\left( {{\rm{aspirin}}} \right)\;{\rm{ = }}\;{\rm{72}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{aspirin}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}$

Using the molar mass of aspirin, we can calculate its mass for 100%-yield reaction:

${\rm{m}}\;{\rm{(aspirin)}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{180}}{\rm{.2}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{13,046}}\;{\rm{g}}$

This corresponds to 13.05 kg when divided by 1000.

d) What mass of aspirin is formed if the reaction yield is 70.0%?

To find the mass of aspirin when the percent yield is 70.0%, we need to multiply the theoretical yield (13.05 kg) by 0.700:

m (aspirin) = 13.05 kg x 0.700 = 9.14 kg

e) If the actual yield of aspirin is 11.2 kg, what is the percent yield?

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

${\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{11}}{\rm{.2}}\;{\rm{kg}}}}{{{\rm{13}}{\rm{.05}}\;{\rm{kg}}}}\;{\rm{ \times }}\;{\rm{100\% }}\;{\rm{ = }}\;{\rm{85}}{\rm{.8}}\;{\rm{\% }}$

f) How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?

First, calculate the moles of 20.0 kg aspirin which is 2.00 x 104 g:

${\rm{n}}\;{\rm{(aspirin)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.0}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{180}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{111}}\;{\rm{mol}}$

According to the mole ratio of the chemical equation, this is how much salicylic acid would be needed because of the 1:1 mole ratio.

However, because the reaction has an 85.0% percent yield, we need to multiply the moles by a factor of

${\rm{n}}\;{\rm{(salicylic acid)}}\;{\rm{ = }}\;{\rm{111}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{131}}\;{\rm{mol}}$

The mass of salicylic acid would be:

${\rm{m}}\;{\rm{(salicylic acid)}}\;{\rm{ = }}\;{\rm{131}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{138}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{18,078}}\;{\rm{g}}$

This corresponds to 1.81 kg

### 4 thoughts on “Reaction/Percent Yield”

1. Thank you so much! This really helped me a lot and especially the practice problems were helpful to understand the concept.