Question

Liquid benzene $(C_6H_6)$ bums in oxygen according to the equation, $ \, \, \, \, \, \, 2C_6H_6 (I) + 15O_2 (g) \rightarrow 12 CO_2 (g) + 6H_2O(g) $ How many litres of $O_2$ at STP are needed to complete the combustion of 39 g of liquid benzene? $(Mol. \, weight \, of \, O_2 = 32, C_6H_6 = 75)$

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

Answer 1

The Correct Option is D

(d) $2C_6 H_6 + 15O _2 (g) \rightarrow 12CO_2 (g) + 6H_2O(g)$
$_{=156}^{2 \times 78} \, \, \, \, \, _{=330}^{15 \times 32} $
$\therefore$ 156 g of benzene required oxygen $= 15 \times 22.4 L $
$\therefore$ 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