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=(Actual)/(Theoretical)=(6g)/(8.09g)=0.7417\to74.17`

Therefore, the percent yield is 74.17%.