In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio which corresponds to three times as much oxygen as is required theorectically for complete combusion of the fuel A burner which has been adjused for methane as fuel (with `x L h^(-1)` of `CH_(4)` and `6x Lh^(-1) of CO_(2))` is to be readjusted for butane `C_(4)H_(10)` in order to get the same calorific output what should be the rate of supply to butane and oxygen? Assume that losses due to incomplete combustion etc are the same for both fuels and that the gases behave ideally Heats of combusion
`CH_(4) =809 kJ mol^(-1),C_(4)H_(10) =2878 kJmol^(-1)` .

Answer

5.48 (X) litre `O_(2)`

Explanation

Combustion of `CH_(4)` and `C_(4)H_(10)` takes places as
`._("Initial volume (inlitre)") underset(X)(CH_(4)) + underset(6 X) (2O_(2)) rarr CO_(2) + 2 H_(2)O`,
`Delta H = - 809 kJ mol^(-1)` Let the temperature be T and assume volume of 1 mle of a gas is V litre at this condition `:.` V litre of 1 mole of `CH_(4)` gives energy on combustion = 809 kJ `:.` X litre of `CH_(4)` gives energy on combustion `= (808(X))/(V) kJ :. 2878 kJ` energy is obtained by 1 mole of V litre `C_(4)H_(10) .: (809(X))/(V) kJ`energy is obtained `.: (809 (X) xxV)/(V xx 2878)` litre`C_(4)H_(10) = 0.281 (X)` litre `C_(4)H_(10)` Thus, butane sullpied for the same calorific output = 0.281 (X) litre
`:. C_(4)H_(10) + (13)/(2)O_(2) rarr 4 CO_(2) + 5H_(2) Ol Delta H = - 2878 kJ//mol`
Volume of `O_(2)` required = 3 volume of `O_(2)` for combustion of `C_(4)H_(10)`
`= 3 xx (13)/(2) xx` volume of `C_(4)H_(10)`
`= 3 xx (13)/(2) xx 0.281 (X)` litre `O_(2)`