Question

Propane burns in air according to the following equation:
$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$.
What volume of propane is consumed on using $1000c{m}^3$ of air, considering only $20$% of air contains oxygen?

Answer 1

Propane burns in air according to the following equation:
$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$.
1 mole of propane combines with 5 moles of oxygen.
Volume is directly proportional to the number of moles.
1 $c{m}^3$ of propane will combine with 5 $c{m}^3$ of oxygen.
Only $20$% of air contains oxygen. So $1000c{m}^3$ of air contains $1000c{m}^3\times \frac{20}{100}=200c{m}^3$ of oxygen.
The volume of propane that is consumed on using $1000c{m}^3$ of air is
$\frac{200c{m}^3}{5}=40c{m}^3$.