Question

Calculate the weight of iron which will be converted to its oxide by the action of $18\text{g}$ of steam on it when the yield of reaction is 50 %.
$2Fe+3H_{2}O\to F{e}_2{O}_3+3H_2$
(Molar mass = $56\text{g/mol}$)

• A. $30.55\text{g}$
• B. $18.76\text{g}$
• C. $24.46\text{g}$
• D. $14.29\text{g}$

Answer 1

The correct option is B $18.76\text{g}$

$2Fe+3H_{2}O\to F{e}_2{O}_3+3H_2$
As per the stoichiometry of the reaction:
Number of moles of $H_{2}O=\frac{18}{18}=1\text{mol}$
3 mol of $H_{2}O$ is required to react with 2 moles of $Fe$
1 mol of $H_{2}O$ will require $\frac{2}{3}$ mol of $Fe$ = 0.67 mol (Theoretical)
Since, the yield of the above reaction is 50 %
Number of moles of Fe converted (actual) = 0.67 $\times$0.5 = 0.335 mol
Amount of $Fe$ converted = number of moles of Fe $\times$ molar mass of Fe
Amount of $Fe=0.335\times 56=18.76\text{g}$