Answer
$V(CH_{4})=415.8l$
Work Step by Step
$T=300K$
$p=825torr=825\times 133.3Pa=109972.5Pa=109.9725kPa$
$m(C_{6}H_{12}O_{6})=1.1kg=1100g$
---$V(CH_{4})=?$
$n(C_{6}H_{12}O_{6})=\frac{m(C_{6}H_{12}O_{6})}{M(C_{6}H_{12}O_{6})}=\frac{1100g}{180\frac{g}{mol}}=6.11mol$
Bacterial breakdown of a simple sugar can be represented by the following reaction equation:
$C_{6}H_{12}O_{6}\rightarrow 3CH_{4}+3CO_{2}$
Since $1mol$ of a simple sugar produces $3mol$ of methane, $n(CH_{4})=3n(C_{6}H_{12}O_{6})=18.33mol$
$pV(CH_{4})=n(CH_{4})RT \implies V(CH_{4})=\frac{n(CH_{4})RT}{p}=\frac{18.33mol\times 8.314\frac{J}{K\times mol}\times 300K}{109.9725kPa}=415.8dm^{3}=415.8l$
