Question

$C_6H_6(\text{liquid}) + \frac{15}{2} O_2(\text{gas}) \rightarrow 6 CO_2(\text{gas}) + 3 H_2O(\text{liquid})$ Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at STP) needed for complete combustion of $39$ grams of liquid benzene?

• A. 11.2 litre
• B. 22.4 litre
• C. 84 litre
• D. 168 litre

Answer 1

The Correct Option is C

The balanced chemical equation for the combustion of benzene is:

$C_6H_6(\text{liquid}) + \dfrac{15}{2} O_2(\text{gas}) \rightarrow 6 CO_2(\text{gas}) + 3 H_2O(\text{liquid})$ 

This tells us that 1 mole of benzene reacts with $\dfrac{15}{2} = 7.5$ moles of oxygen. 

Molar mass of benzene, $C_6H_6$: 
$6 \times 12 + 6 \times 1 = 72 + 6 = \mathbf{78\ g/mol}$ 

Given mass of benzene = 39 g
Moles of benzene = $\dfrac{39}{78} = \mathbf{0.5\ mol}$ 

Oxygen required = $0.5 \times 7.5 = \mathbf{3.75\ mol}$ 

At STP, 1 mol of a gas occupies 22.4 L
Volume of oxygen needed = $3.75 \times 22.4 = \mathbf{84\ L}$ 

Answer: 84 litre