A natural gas may be assumed to be a mixture of methane and ethane only. Complete combustion of 10 L of gas at S.T.P. the heat evolved was 474.6 KJ. Assuming (ΔCH)CH4(g)=−894KJ.mol−1 and (ΔCH)C2H6(g)=−1500 KJ.mol−1 .Calculate the percentage composition of the mixture by volume is around

Detailed Solution: Let us consider the combustion processes of methane and ethane.CH4+2O2→CO2+2H2OC2H6+72O2→2CO2+3H2OLet us consider the followingnTotal =1022.41=0.44 mol;nCH4=n1;nC2H6=n2;n1+n2=0.44-894×n1+(-1500)×n2=-474.6n1+1.68n2=0.53   (2)  (2) - (1) 0.68n2=0.53-0.44n2=0.133The percentage of methane in the mixture is 0.310.44×100=70.45And the remaining is ethane.