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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» Correct Answer - 2 L, 1L 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`. |
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