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A natural gas may be assumed to be a mixture of CH_4 and C_2H_6 only. On complete combustion of 10 L of the gas at STP, the heat evolved was 474.6 kJ. Assuming DeltaHc(CH_4) = – 894 kJ//mol and ""DeltaHC(C_2H_6) = - 1560 kJ//"mole". Calculate the % by volume of each gas in the mixture

Answer»

Solution :LET, V be the VOLUME of `CH_4` in 10L MIXTURE. We have
Amount of methane = `v/(22.4)`, Amount of ethane = `(10- v)/(22.4)`
The expression of heat evolved, we will get
`V/(22.4) xx 894 + (10 - V)/(22.4) xx 1560 = 474.6`
V = 7.45 litre
% of `CH_4 = 74.5%`
% of `C_2H_6 = 25.5%`.


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