Q=mc∆T=15000 x 4.186 x 76.6 = 4 809 714 J.
ΔHcºCH4 = (1*ΔHfºCO2 + 2*ΔHfºH2O) -(1*ΔHfºCH4 + 2*ΔHfºO2)
= (1*-394 kJ/mol + 2*-293 kJ/mol) – (1*-74.5 kJ/mol + 2*0kJ/mol)
=-905.5 kJ/mol
4 809 714 J./0.129 x 1000 = 37 284, 6047
37 284,6047/905.5= 41,1757092 moles of CH4.
16 x 41,1757092/0.668 = 986,244531 L.
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