What will be the amount of heat evolved by burning 10 L of methane under standard conditions?
`-76.2, -398.8 and -241.6" kJ mol"^(-1)` resepctively

Options

A. 805.8kJ
B. 398.8 kJ
C. 359.7 kJ
D. 640.4 kJ

Answer

C

Explanation

`CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O`
`DeltaH=H_(p)-H_(R)`
`=DeltaH_(f)^(@)(CO_(2))+2DeltaH_(f)^(@)(H_(2)O)-[DeltaH_(f)^(@)(CH_(4))+2DeltaH_(f)^(@)(O_(2))]`
`=-398.8-2xx241.6-(-76.2+2xx0)=-"805.8 kJ mol"^(-1)`
`"22.4 L of "CH_(4)" (1 mole ) gives 805.8 kJ"`
`"10 L of "CH_(4)" will give "(805.8)/(22.4)xx10=359.73kJ`