Question

Liquid benzene $\left(C_6{H}_6\right)$ burns in oxygen according to the following equation:
$2C_6{H}_6\left(l\right)+15O_2\left(g\right)\to 12C{O}_2\left(g\right)+6H_{2}O\left(g\right)$
How many litres of oxygen at STP are required for the complete combustion of $39\text{g}$ of liquid benzene?

• A. $74\text{L}$
• B. $11.2\text{L}$
• C. $22.4\text{L}$
• D. $84\text{L}$

Answer 1

The correct option is D $84\text{L}$

$\text{Moles of benzene}=\frac{\text{given mass}}{\text{molar mass}}=\frac{39}{78}=0.5\text{mol}$

The given balanced chemical reaction is:
$2C_6{H}_6\left(l\right)+15O_2\left(g\right)\to 12C{O}_2\left(g\right)+6H_{2}O\left(g\right)$

As per the reaction, $2$ moles of benzene reacts with $15$ moles of $O_2$.

Thus, $2$ moles of benzene reacts with $(15\times 22.4)\text{L}$ of $O_2$.
Hence, $0.5$ moles of benzene will react with $\left(\frac{15\times 22.4}{2}\right)\times 0.5\text{L of}{O}_2$
$=84\text{L of}{O}_2$