Question

The balanced equation for the preparation of $NaBr$ is as follows:
$Fe+B{r}_2\to FeB{r}_2\text{…(i) (yield: 100 \%)}$
$3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8\text{...(ii) (yield: 100 \%)}$
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4\text{…(iii) (yield: 60 \%)}$
Calculate the mass of iron required to produce $2.06\times {10}^3\text{g}$ of $NaBr$.
(Considering $N{a}_{2}C{O}_3$ and $B{r}_2$ to be present in excess)
(molar mass of $Fe=56\text{g/mol},Br=80\text{g/mol},Na=23\text{g/mol}$)

• A. $420\text{g}$
• B. $4200\text{g}$
• C. $251\text{g}$
• D. $700\text{g}$

Answer 1

The correct option is D $700\text{g}$

The balanced chemical equations for the preparation of $NaBr$ are as follows:
$3Fe+3B{r}_2\to 3FeB{r}_2\text{---(i)}$
$3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8\text{----(ii)}$
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4\text{---(iii)}$
From the balanced chemical equations, it can be seen that 3 mol of $Fe$ produces 3 mol of $FeB{r}_2$ which in turn will produce 1 mol of $F{e}_{3}B{r}_8$.
$\Rightarrow$ Number of moles of $Fe$ required = 3$\times$ number of mol of $F{e}_{3}B{r}_8$ required
The molar masses of $F{e}_{3}B{r}_8$ and $NaBr$ are $808\text{g/mol}$ and $103\text{g/mol}$ respectively.
Since, 8 mol of $NaBr$ is produced from 1 mol of $F{e}_{3}B{r}_8$
$(8\times 103)\text{g}$ of $NaBr$ is produced from $(1\times 808)$ of $F{e}_{3}B{r}_8$
$2.06\times {10}^3\text{g}NaBr$ will be produced by $=\frac{1\times 808}{8\times 103}\times 2.06\times {10}^3\text{g}$ of $F{e}_{3}B{r}_8$
Mass of $F{e}_{3}B{r}_8$ required = $2020\text{g}$
Since, the yield of the third reaction is 60%, then the actual amount of $F{e}_{3}B{r}_8$ required will be:
Amount of $F{e}_{3}B{r}_8$ (required) = $\frac{2020}{0.6}=3366.67\text{g}$
$\Rightarrow$ Amount of $Fe$ (required)= $=\frac{3\times 56\times 3366.67}{808}=700\text{g}$
Therefore, the actual amount of Fe required will be equal to $700\text{g}$.