Answer
(a) $mass = 5.04 \times 10^{2} \ kg$
(b) $heat = -6.20 \times 10^{5} \ kJ$
(c) 2.9 L of water
Work Step by Step
(a)
$ CH_4 $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )= 16.04 g/mol
1. Calculate the mass.
$$mass = -2.80 \times 10^{7} \ kJ \times \frac{ 1.00 \ mol \ CH_4 }{ -890.3 \ kJ} \times \frac{ 16.04 \ g \ CH_4 }{1 \ mol \ CH_4 } $$
$$mass = 5.04 \times 10^{5} \ g = 5.04 \times 10^{2} \ kg$$
(b)
$$Pressure = 768 \space mmHg \times \frac{1 \space atm}{760 \space mmHg} = 1.01 \space atm$$
$$Temperature/K = 18.6 + 273.15 = 291.8 \space K$$
1. According to the Ideal Gas Law:
$$n = \frac{PV}{RT} = \frac{( 1.01 \space atm )( 1.65 \times 10^{4} \space L )}{( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 291.8 \space K)}$$
$$n = 696 \space mol$$
2. Calculate the resulting heat.
$$heat = 696 \ mol \times \frac{ -890.3 \ kJ}{ 1.00 \ mol \ CH_4 }$$
$$heat = -6.20 \times 10^{5} \ kJ$$
(c)
q = m * specific heat * $\Delta T$
$$m = \frac{q}{sh*\Delta T} $$
$$m = \frac{ 6.20 \times 10^{5} }{ 4.18 * ( 60.0 - 8.8 ) } $$
$m = 2.9 \times 10^{3} \ g = 2.9 \ kg$
At this temperature range, thee density of water is approximately constant and equal to 1.0 g/mL or 1.0 kg/L.
$V = 2.9 \ kg \times \frac{1 \ L}{1.0 \ kg} = 2.9 \ L$