Question

$12.6g$ nonene $C_9{H}_{18}$ is combusted in the presence of air. The percentage of oxygen in air is 21 percent by volume at STP. The volume of air needed for complete combustion of nonene is

• A. $112L$
• B. $144L$
• C. $22.4L$
• D. $224L$

Answer 1

The correct option is B $144L$

Balanced equation is
$C_9{H}_{18}+\frac{27}{2}{O}_2\to 9C{O}_2+9H_{2}O$
Given: Mass of nonene = 12.6 g
Number of moles of nonene $=\frac{12.6}{126}=0.1mol$
Moles of $O_2$ at STP $=\frac{\frac{27}{2}}{1}\times 0.1$
$=\frac{2.7}{2}mol$
Volume of $O_2$ at STP $=\frac{2.7}{2}\times 22.4=2.7\times 11.2$ L
$21L$ of oxygen is present $100L$ of air
volume of air required $=\frac{100}{21}\times 2.7\times 11.2=144L$