Question

If yield of reaction (iii) is $90\%$, then kilomoles of $C{O}_2$ formed when $2.06\times {10}^3$ kg $NaBr$ formed is :

• A. $20$
• B. $10$
• C. $40$
• D. none of the above

Answer 1

The correct option is B $10$

The molar mass of $NaBr$ is $103$ g/mol.
$2.06\times {10}^3$ kg $NaBr$ corresponds to $\frac{2.06\times {10}^6}{103}=20000$ moles of $NaBr$.
$8$ moles of $NaBr$ corresponds to $4$ moles of $C{O}_2$.
Hence, $20000$ moles of $NaBr$ corresponds to $10000$ moles of $C{O}_2$.