Question

Volume of oxygen required for complete combustion of 12.5 $d{m}^3$ of benzene at NTP is:

• A. 87.5 $d{m}^3$
• B. 43.75 $d{m}^3$
• C. 40 $d{m}^3$
• D. 80 $d{m}^3$

Answer 1

The correct option is A 87.5 $d{m}^3$

The chemical reaction can be represented as:

$C_6{H}_6+7.5O_2\to 6C{O}_2+3H_{2}O$

Since at NTP, 1 mole=22.4 litres, hence

According to the reaction, $22.4L$ benzene requires $7.5\times 22.4=168L$ oxygen for complete combustion

Therefore, $12.5d{m}^3$ benzene will require $\frac{168}{22.4}\times 12.5=87.5d{m}^3$ oxygen for complete combustion

Hence, option $A$ is the correct answer.