Question

If the yield of second reaction is $60$ percent and third reaction is $70$ percent then, mass of iron required to produce $2.06\times {10}^3$ kg $NaBr$ is :

• A. ${10}^5$ kg
• B. ${10}^5$ g
• C. ${10}^3$ kg
• D. None

Answer 1

Answer: (C)

${10}^3$ kg

The balanced chemcial equations for the preparation of NaBr are as follows:

$Fe+B{r}_2\to FeB{r}_2$

$3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8$

$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4$

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 .

$\Rightarrow 2.06\times {10}^3$ Kg NaBr $\times \frac{3\times 55.8}{8\times 103}\approx$ 420 Kg Fe .

If the yield of reaction (ii) is 60% and reaction (iii) is 70%. then mass of iron required to produce $2.06\times {10}^3$ Kg NaBr will be $420\times \frac{100}{60}\times \frac{100}{70}\approx {10}^{3}Kg$

Hence, option C is correct.