Question

Liquid benzene $(C_6{H}_6)$ bums in oxygen according to the equation,
$2C_6{H}_6(I)+15O_2(g)\to 12C{O}_2(g)+6H_{2}O(g)$
How many litres of $O_2$ at STP are needed to complete the combustion of 39 g of liquid benzene?
$(Mol.weightof{O}_2=32,C_6{H}_6=75)$

• A. 74 L
• B. 11.2 L
• C. 22.4 L
• D. 84 L

Answer 1

Answer: (D)

(d) $2C_6{H}_6+15O_2(g)\to 12C{O}_2(g)+6H_{2}O(g)$
${}_{=156}^{2\times 78}{}_{=330}^{15\times 32}$
$\therefore$ 156 g of benzene required oxygen $=15\times 22.4L$
$\therefore$ 1 g of benzene required oxygen $=\frac{15\times 22.4}{156}L$
$\therefore$ 39 g of benzene required oxygen $=\frac{15\times 22.4\times 39}{156}$
= 84.0 L