Question

What volume of oxygen gas at NTP is necessary for complete combustion of 20 litre of propane measured at ${27}^{o}C$ and 760 mm pressure ?

Answer 1

The balanced equation is:
$\underset{1litre}{\underset{1vol.}{C_3{H}_8}}+\underset{5litre}{\underset{5vol.}{5O_2}}\to 3C{O}_2+4H_{2}O$

1 litre of propane requires = 5 litre of oxygen
20 litre of propane will require$=5\times 20=100litre$ of oxygen at 760 nun pressure and ${27}^{o}C$.
This volume will be converted to NTP conditions.
Given conditions
$P_1=760mm$
$V_1=100litre$
$T_1=27+273=300K$
NTP conditions
$P_2=760mm$
$V_2=?$
$T_2=273$

Applying :

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$V_2=\frac{P_1{V}_1}{T_1}\times \frac{T_2}{P_2}$

$=\frac{760\times 100}{300}\times \frac{273}{760}=91.0litre$