Question

3. Combining 167.7 g of Fe with excess water produces 6.00 g of hydrogen gas. What is
the percent yield?

Answer 1

The balanced chemical equation is

$3Fe(s)+4H_2O(l)\to Fe_3O_4(s)+4H_2(g)$

We know that the reaction produces 6 grams of hydrogen gas, this is the actual yield. We need to find the theoretical yield using the stoichiometry of the chemical equation.

According to the equation, 3 moles of Fe produces 4 moles of hydrogen gas, this is the ratio, we know that the molar mass of Fe is 55.845 gr/mol.

$167.7g(Fe)\cdot\frac{1\text{mol(Fe)}}{55.845g(Fe)}\cdot\frac{4moles(H_2)}{3\text{moles(Fe)}}\cdot\frac{2.02g(H_2)}{1\text{mole(H}_2)}=8.09g(H_2)$

The theoretical yield is 8.09 grams of hydrogen gas. Now we can find the percent yield.

$Yield=\frac{Actual}{Theoretical}=\frac{6g}{8.09g}=0.7417\to74.17$

Therefore, the percent yield is 74.17%.