Question

$NaBr$, used to produce $AgBr$ for use in photography can itself be prepared as follows:
$Fe+B{r}_2\to FeB{r}_2$
$FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8$ (not balanced)
$F{e}_{3}B{r}_8+N{a}_{2}C{O}_3\to NaBr+C{O}_2+F{e}_3{O}_4$ (not balanced)
How much $Fe$, in kg, is consumed to produce $4.12\times {10}^{3}kgNaBr$

Answer 1

On converting the given amount of $NaBr$ into mole $=\frac{W}{M}=\frac{4.12\times {10}^3\times {10}^{3}g}{(23+80)}=\frac{4.12\times {10}^6}{103}mol$
Balanced reaction
$\left(i\right)(Fe+B{r}_2\to FeB{r}_2)\times 3$
$\Rightarrow 3Fe+3B{r}_2\to 3FeB{r}_2$
$\left(ii\right)3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8$
$\left(iii\right)F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4$
From stoichiometry of all the reaction 8 mole of $NaBr$ is obtained from 1 mole of $F{e}_{3}B{r}_8.$ Which is obtained from 3 mole of $FeB{r}_2$ in previous reaction.
According to first reaction 3 mole of $FeB{r}_2$ is obtained from 3 mole of $Fe$. Hence, we can say that 8 mole of $NaBr$ will be obtained from 3 mole of $Fe$.
$\because$ 8 mole of $NaBr$ is obtained from = 3 mole of $Fe$
$\therefore \frac{4.12\times {10}^6}{103}$ mole of $NaBr$ will be obtained from $=\frac{3}{8}\times \frac{4.12\times {10}^6}{103}$ mole of $Fe$
$\therefore$ wt. of required $Fe=n\times molarmass$
$\frac{3}{8}\times \frac{4.12\times {10}^6}{103}\times 56=840kg$