Question

When \(160\,\mathrm{g}\) of methane was burnt in air at STP, \(5\%\) of it remained. At \(273\,\mathrm{K}\) and \(1\,\mathrm{bar}\) pressure, air contains \(20\%\) of \(\mathrm{O_2}\) by volume. What is the approximate volume (in L) of air consumed? \[ \text{(At STP molar volume }=22.71\,\mathrm{L}) \]

• A. \(2157.45\)
• B. \(215.74\)
• C. \(21574.5\)
• D. \(431.49\)

Hint:

For combustion of methane, \[ \boxed{ \mathrm{CH_4+2O_2\rightarrow CO_2+2H_2O} } \] If air contains \(20\%\) oxygen, \[ \boxed{ V_{\text{air}}=\frac{V_{O_2}}{0.20}. } \]

Answer 1

The Correct Option is A

Step 1: Calculate the moles of methane burnt. \[ \mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O} \] Initial moles of methane, \[ n=\frac{160}{16}=10\ \text{mol}. \] Since \(5\%\) remains unburnt, \[ 95\% \text{ burns}. \] Hence, \[ n_{\text{burnt}} =0.95\times10 =9.5\ \text{mol}. \]

Step 2: Calculate oxygen required. From the reaction, \[ 1\ \text{mol CH}_4 \rightarrow 2\ \text{mol O}_2. \] Therefore, \[ n_{O_2} = 2\times9.5 = 19\ \text{mol}. \]

Step 3: Calculate the volume of air consumed. Volume of oxygen, \[ V_{O_2} = 19\times22.71 = 431.49\ \mathrm{L}. \] Since air contains \(20\%\) oxygen, \[ V_{\text{air}} = \frac{431.49}{0.20} = 2157.45\ \mathrm{L}. \] Hence, \[ \boxed{2157.45\ \mathrm{L}} \] Therefore, \[ \boxed{(A)} \] is the correct answer.