Question

Mass of iron required to produce $2.06\times {10}^3$kg $NaBr$ is:
[Atomic weight of $Br=80,Fe=56$]

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

Answer 1

The correct option is B $420$ kg

The balanced chemical equations are as given below:
$Fe+B{r}_2\to FeB{r}_2$
$3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8$
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4$
Thus, $3$ moles of iron produces $8$ moles of $NaBr$.
The molar masses of $Fe$ and $NaBr$ are $55.8$ g/mol and $102.9$ g/mol respectively.
The mass of iron required to produce $2.06\times {10}^3$ $NaBr$ is $2.06\times {10}^3\times \frac{3\times 55.8}{8\times 102.9}=418.9$ kg $\simeq 420$ kg.