Stoichiometric air; from the reaction equation $1$mol methane requires $2$mol oxygen,

$so,$ stoichiometric air $=9\mathrm{.}790\mathrm{.}952\frac{100}{21}=88\mathrm{.}6(mol{)}_{{?}_{?}}$

Percentage excess air $=\frac{(\text{a}\text{i}\text{r}\text{s}\text{u}\text{p}\text{p}\text{l}\text{i}\text{e}\text{d}-\text{s}\text{t}\text{o}\text{i}\text{c}\text{h}\text{i}\text{o}\text{m}\text{e}\text{t}\text{r}\text{i}\text{c}\text{a}\text{i}\text{r})}{\text{s}\text{t}\text{o}\text{i}\text{c}\text{h}\text{i}\text{o}\text{m}\text{e}\text{t}\text{r}\text{i}\text{c}\text{a}\text{i}\text{r}}100$

$=\frac{108\mathrm{.}4-88\mathrm{.}6}{88\mathrm{.}6}=22$ per cent

$1\mathrm{.}4.$ To ensure complete combustion, $20\%$ excess air is supplied to a furnace burning natural gas. The gas composition $($by volume$)$ is methane $95\%,$ ethane $5\%.$ Calculate the moles of air required per mole of fuel.

Solution

Basis: $100$mol gas, as the analysis is volume percentage.

Reactions: $C{H}_4+2O_{2}C{O}_2+2H_{2}O$

$C_2{H}_6+3\frac{1}{2}{O}_{2}2C{O}_2+3H_{2}O$

Stoichiometric mols $O_2$ required $=952+53\frac{1}{2}={207\mathrm{.}5}_{{?}_{?}}$

With $20$ per cent excess, mols $O_2$ required $=207\mathrm{.}5\frac{120}{100}={249}_{{?}_{?}}$

Mols air $(21$ per cent $O_2)=249\frac{100}{21}=1185\mathrm{.}7$

Air per mol fuel $=\frac{1185\mathrm{.}7}{100}=11\mathrm{.}86(mol{)}_{{?}_{?}}$

$31$