Question

Calculate the volume of air containing 21% by volume of oxygen at NTP required to convert 294 $c{m}^3$, of $S{O}_2$ into $S{O}_3$, under the same conditions.

• A. 200 ml
• B. 300 ml
• C. 500 ml
• D. 700 ml

Answer 1

The correct option is D 700 ml

$S{O}_2+\frac{1}{2}{O}_2⟶S{O}_3$

At NTP $22.4L\frac{22.4}{2}L$

$\because 22.4L$of $S{O}_2$ requires $11.2L$ of O_2,$

$\therefore 294c{m}^3$ of $S{O}_2$ requires $\frac{11.2}{22.4}\times 294=147c{m}^3$ of $O_2.$

As air contains $21\%$ by volume of oxygen, at NTP,

So, $\frac{147}{21}\times 100=700ml$