Collect and Organize This is a stoichiometry problem like Sample Exercises 3.12 and 3.13, except that the yield of reaction is only 96%.

Analyze We need to convert the mass of H_{2} we wish to make into moles of H_{2}, and then we use the mole ratios of H_{2}:CO and H_{2}:H_{2}O to find the amount of reactants we need. To produce 176 g of H_{2} by using a reaction that produces 96% of the theoretical yield, we consider the theoretical yield to be 96% of the total yield we need. Then by analogy to Sample Exercise 3.13, we can calculate the amount of both reactants needed.

Solve
1. The number of moles of H_{2} we wish to produce is:

176  \sout{g  H_{2}} \times \frac{1  mol  H_{2}}{2.016  \sout{g  H_{2}}} =87.3  mol  H_{2}

However, since the reaction produces only 96% of the theoretical yield, we need to base our calculation on what 100% yield would be if 87.30 moles is 96% of the theoretical yield:

87.30  mol = 0.96x             x = 91  mol

2. Using the mole ratios of H_{2}:H_{2}O and H_{2}:CO, we find the number of moles of H_{2}O and CO:

91  \sout{mol  H_{2}} \times \frac{1  mol  H_{2}O}{1  \sout{mol  H_{2}}} =91  mol  H_{2}O                91  \sout{mol  H_{2}} \times \frac{1  mol  CO}{1  \sout{mol  H_{2}}} =91  mol  CO

3. We convert from moles of reactants to the mass of the reactants by multiplying by the molar mass:

91  \sout{mol  H_{2}O} \times \frac{18.02  g  H_{2}O}{1  \sout{mol  H_{2}O}} =1.6 \times 10^{3}  g  H_{2}O

91  \sout{mol  CO} \times \frac{28.01  g  CO}{1  \sout{mol  CO}}= 2.5 \times 10^{3}  g  CO

As a check of our answers, let’s calculate how much H_{2} would be produced from 91 moles of CO if the reaction proceeded in 100% yield:

91  \sout{mol  CO} \times \frac{1  \sout{mol  H_{2}}}{1  \sout{mol  CO}} \times \frac{2.016  g  H_{2}}{1  \sout{mol  H_{2}}} =183  g  H_{2}

The percent yield is

\frac{176  g  H_{2}}{183  g  H_{2}} \times 100\% = 96\%

Think About It On a large scale, any yield less than 100% could be a significant cost for rare and expensive reactants.