Answer
$-3.55*10^{4}\text{ kJ}$
Work Step by Step
a. $1.00 \text{ gram $CH_4$}* \frac{1\text{ mol $CH_4$}}{16.04\text{ grams $CH_4$}}*\frac{-891\text{ kJ}}{1\text{ mol $CH_4$}}=-55.5\text{ kJ}$
b.
$PV=nRT$
$740*1.00*10^{3}=n*62.364*298$
$n=39.8\text{ mol}$
$39.8\text{ mol}*\frac{-891\text{ kJ}}{1\text{ mol}}=-3.55*10^{4}\text{ kJ}$
