CH4 (g) + 2 O2 (g) ==> CO2 (g) + 2 H2O (l) ; initially water in liquid state only, but later due to high temperature it is gaseous
∆H0rxn = [-393.5 + (2 x -285.80] - [-74.6 + (2 x 0)] = -890.5 kJ/mol
amount of heat required to heat 18000 g of water from 23.2 0C to 100 0C
Q = m x Cs x ΔT
m = mass; Cs = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released.
To heat the water from 23.2 °C to boiling at 100°C
= 18000 g x 4.184 J/g/°C x 76.8°C = 5783.9616 kJ
To vaporize the water to steam (gaseous) at 100°C
= 18000 g x 2,260 J/g = 40680000 J = 40680 kJ.
Total heat = 5783.9616 kJ + 40680 kJ = 46463.9616 kJ
but efficiency = 10.4 %
heat required will be = 46463.9616 kJ / 10.4% = 446768.861538 kJ required
mol methane required = 446768.861538 / 890.5 kJ/mol = 502 501.70562778 mol
mass of methane = 501.70562778 mol x 16.04 g/mol = 8047.3582696 g
volume = 8047.3582696 g / 0.660 g/L = 12192.9670751 L
volume of methane = 12193 L
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