When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0^@C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat released from this combustion in kJ is (DeltaH_C(CH_4)=-890 kJ mol^(-1) and DeltaH_C(C_3H_8)=-2220 kJ "mol"^(-1))

When 15.68 litres of a gas mixture of methane and propane are fully co

1.	When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0^@C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat released from this combustion in kJ is (DeltaH_C(CH_4)=-890 kJ mol^(-1) and DeltaH_C(C_3H_8)=-2220 kJ "mol"^(-1))
Answer» `-889 kJ` `-1390` kJ `-3180` kJ `653.66`kJ Solution :Given: `DeltaH_C(CH_4)=-890 "kJ mol"^(-1)` `DeltaH_C(C_3H_8)=-"-2220 kJ mol"^(-1)` Let the mixture contain X lit of and lit of propane . `undersetx(CH_4)+underset(2X)(2O_2) to CO_2 + 2H_2O` `underset(15.68-x)(C_3H_8+5O_2) to underset"5(15.68-x)"(3CO_2 + 4H_2O)` Volume of oxygen consumed = 2x+5 (15.68-x)=32 lit 2x + 78.4 -5x =32 78.4 - 3X =32 3x=46.4 x=15.47 Given mixture contains 15.47 liters of methane and 0.213 liters of propane . Hence , `DeltaH_C=[((DeltaH_C(CH_4))/"22.4 lit")xx"lit"]+[((DeltaH_C(C_3H_8))/"22.4 lit")(15.68-x)"lit"]` `DeltaH_C=[((-890 "kJ mol"^(-1))/"22.4 lit")xx"15.47 lit "]+[((-2220)/"22.4 lit")"0.21 lit"]` `DeltaH_C=[-614.66 "kJ mol"^(-1)]+[-20.81 "kJ mol"^(-1)]` `DeltaH_C=635.47 "kJ mol"^(-1)`