Question

Under constant temperature and pressure $50\text{ml}$ of gaseous mixture of acetylene and ethylene if taken in a ratio of $a:b$ requires $700\text{ml}$ of air containing $20\%$ by volume $O_2$ for complete combustion. Calculate the volume of air required for complete combustion of a mixture having ratio $b:a$ at same conditions of temperature and pressure.

• A. $700\text{ml}$
• B. $675\text{ml}$
• C. $135\text{ml}$
• D. $140\text{ml}$

Answer 1

The correct option is B $675\text{ml}$

Acetylene and ethylene are taken in a ratio of $a:b$.
Acetylene $\underset{a}{C_2{H}_2}+\underset{5a/2}{\frac{5}{2}{O}_2}\to 2C{O}_2+H_{2}O$
Ethylene $\underset{b}{C_2{H}_4}+\underset{3b}{{30}_2}\to 2C{O}_2+2H_{2}O$
This mixture requires $700\text{ml}$ of air containing $20\%$ by volume $O_2$ for complete combustion.
$\therefore \frac{5}{2}a+3b=700\times 0.2........\left(1\right)$
$a+b=50........\left(2\right)$
Solving equation $1$ and $2$, we get
$a=20,b=30$
If the ratio of acetylene and ethylene is $b:a$, then
$\frac{5b}{2}+3a=V\times 0.2$
$\Rightarrow \frac{5}{2}\times 30+3\times 20=V\times 0.2\Rightarrow 135=V\times 0.2\Rightarrow V=675\text{ml}$