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## Q. 3.23

One way of producing hydrogen for use in hydrogen powered vehicles is the water–gas shift reaction:

$H_{2}O(g) + CO(g) → H_{2}(g) + CO_{2}(g)$

At 200°C, the reaction produces a 96% yield. How many grams of $H_{2}O$ and CO are needed to generate 176 g of $H_{2}$ under these conditions?

## Verified Solution

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.