Question

What volume of air at $1$ atm and $273\text{K}$ that has 21% of oxygen by volume is required to completely burn sulphur $\left(S_8\right)$ present in $200\text{g}$ of a sample, which contains 20% inert material that does not burn? Sulphur burns according to the reaction:
$\frac{1}{8}{S}_8\left(s\right)+O_2\left(g\right)\to S{O}_2\left(g\right)$

• A. 23.52 litre
• B. 320.0 litre
• C. 112.0 litre
• D. 533.3 litre

Answer 1

The correct option is D 533.3 litre

1 mol of oxygen is required to burn 1 mol of S.
1 mol of oxygen is required to burn 32g of S
So, to burn 160 g of sulphur $\frac{160}{32}$, 5 mol of $O_2$ is required
At STP, $5\times 22.4L$ of oxygen is consumed in the reaction.
Since 21% of oxygen is consumed in the reaction,
The total volume of air involed in the reaction is =$\frac{100}{21}\times 5\times 22.4L$=533.3 L