Question

Mass of iron required to produce $2.06\times {10}^3$ kg $NaBr$ is :

• A. $420$ g
• B. $420$ kg
• C. $4.2\times {10}^5$ kg
• D. $4.2\times {10}^8$ g

Answer 1

Answer: (B)

$420$ kg

The balanced chemcial equations for the preparation of $NaBr$ are as follows:
$Fe+B{r}_2⟶FeB{r}_2$ .....(i)
$3FeB{r}_2+B{r}_2⟶F{e}_{3}B{r}_8$ .....(ii)
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3⟶8NaBr+4C{O}_2+F{e}_3{O}_4$ .....(iii)
From the balanced chemical equation, it can be seen that $8$ moles of $NaBr$ are produced from $3$ moles of $Fe$.
The molar masses of $Fe$ and $NaBr$ are $55.8$ g/mol and $103$ g/mol respectively.
Thus, mass of $Fe$ in kg is consumed to produce $2.06\times {10}^3$ kg $NaBr$

$=2.06\times {10}^3$ kg $NaBr$ $\times \frac{3\times 55.8}{8\times 103}=420$ kg $Fe$