In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as required theoretically for complete combustion of fuel. A burner which has been adjusted for methane as fuel (with x litre/hour of `CH_(4)` and 6x litre/hour of `O_(2)`) is to be readjusted for butane `C_(4)H_(10)`. In order to get some calorific output, what should be the rate of supply of butane and oxygen? Assume that losses due to incomplete combustion etc., are the same for both fuels and that gases behave ideally. Heals of combustion, `CH_(4) =-809 kJ//"mol", C_(4)H_(10) = -2878` kJ/mol.

Explanation

Calculation of `C_(4)H_(10)`
`CH_(4)+ 2O_(2) rarr CO_(2) + 2H_(2)O, DeltaH = - 809 kJ mol^(-1)`
Initial volume `//` hr (L) x 6x
Suppose the volume of 1 mole of the gas under the given conditions is V litre.
Thus, V litre (1 mole)of `CH_(4)` give energy on combustion `= 809 kJ`
`:. ` x litre of `CH_(4)` will give energy on combustion
`= (809 x)/( V ) kJ`
Now, 2878 kJ of energy is obtained from 1 moleor V litre of `C_(4) H_(10)`
`:. ( 809x)/( V) kJ` of energy will be obtained from `C_(4)H_(10) = ( 809 x xx V )/( V xx 2878) = 0.281 x ` litre
Thus, butane supplied for same calorific output `= 0.281 x` litre `//` hr.
Calculation of`O_(2)`
`C_(4)H_(10) + ( 13)/( 2) O_(2) rarr 4CO_(2) + 5H_(2)O , DeltaH = - 2878 kJ mol^(-1)`
Volume of `O_(2)` required `= 2 xx` volume of `O_(2)` for combustion of `C_(4)H_(10)` (Given )
`= 3 xx ( 13)/(2) xx ` Vol. of `C_(4) H_(10)= 3 xx ( 13)/(2) xx 0.28 x = 5.48 x `litre `//` hr