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The volume of oxygen required for complete combustion of 0.25 mole of methane at S.T.P. is
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$11 \cdot 2 \mathrm{dm}^{3}$
(C)
$\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
1 mole of methane required $=2 \times 22.4 \mathrm{dm}^{3}$ of $\mathrm{O}_{2}$
$\therefore 0.25$ mole of methane required $=2 \times 22.4 \times 0.25=11.2 \mathrm{dm}^{3}$ of $\mathrm{O}_{2}$
$\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
1 mole of methane required $=2 \times 22.4 \mathrm{dm}^{3}$ of $\mathrm{O}_{2}$
$\therefore 0.25$ mole of methane required $=2 \times 22.4 \times 0.25=11.2 \mathrm{dm}^{3}$ of $\mathrm{O}_{2}$
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