A gaseous mixture of 3 L of propane `(C_(3)H_(8))` and butane `(C_(4)H_(10))` on complete combustion at `25^(@)C` produced 10 L of `CO_(2)`. Find out the composition of the gaseous mixture.

Answer

2 L, 1L

Explanation

Let the volume of propane in the mixture = xL
`:.` The volume of butane in the mixture `= (3-x)L`
Now let us calculate the volume of `CO_(2)` evolved with the help of chemical equations
Step I. Calculation of volume of `CO_(2)` from xL of propane
The combustion equation for propane is :
`underset(1L)(C_(3)H_(8))+5O_(2)rarrunderset(3L)(3CO_(2))+4H_(2)O`
1 L of propane `(C_(3)H_(8))` from `CO_(2)=3L`
x L of propane `(C_(3)H_(8))` from `CO_(2)=3xL`
Step II. Calculation of volume of `CO_(2)` from (3-x)L of butane
The combustion equation for butane is :
`underset(1L)(C_(4)H_(10))+(13)/(2)O_(2)rarrunderset(4L)(4CO_(2))+5H_(2)O`
1L of butance `(C_(4)H_(10))` from `CO_(2)=4L`
(3-x)L of butane `(C_(4)H_(10))` from `CO_(2) = 4 xx (3-x)L`
Step III. Calculation of composition of the mixture
Total volume of `CO_(2)` formed in the step I and step II = [3x + 4(3-x)]L
But the volume of `CO_(2)` actually formed = 10 L
`:. 3x +4(3-x)=10 , 3x +12 - 4x =10 - x = -2 or x = 2L`
`:.` Volume of propane = 2L
Volume of butane `= (3-2)=1L`.