Find out the heat evolved in combustion if `112` litres ( at `STP`) of water gas (mixture of eqal volume of `H_(2)(g)` and `CO(g)))`. `{:(H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),,,,DeltaH= -241.8 kJ),(CO(g)+1//2O_(2)(g)rarrCO_(2)(g),,,,DeltaH= -283 kJ):}`

Answer

`1312 kJ`

Explanation

Equal volume of `H_(2)(g) & CO_(g)`
Total volume `=112 L`
So , volume of `CO= "volume of" H_(2)=56 L`
Mole of `CO= "Mole of " H_(2)=2.5 "mole"`
`{:(,H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(g),,,DeltaH= -241.8 kJ),("For",1"mole",,,DeltaH= -241.8 kJ),("For",2.5"mole",,,DeltaH= -241.8xx2.5),("Similarly"",",CO_(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),,,DeltaH= -283 kJ),(,,,,Delta H= -283xx2.5 kJ):}`
Total Heat evolved `=[-241.8+(-283]2.5= -1312 kJ`