The percentage by volume of `C_(3)H_(8)` in a gaseous mixture of `C_(3)H_(8),CH_(4)` and `CO` is 20. When `100mL` of the mixture is burnt in excess of `O_(2)` the volume of `CO_(2)` produced is :

Options

A. `90mL`
B. `160mL`
C. `140 mL`
D. None of these

Answer

3

Explanation

`100 mL` gaseous mixture contain `20 mL C_(3)H_(8)`
So, volume of `CH_(4)` and `CO=(100-20)=80mL`
`C_(3)H_(8)+5O_(2) rarr 3CO_(2)+4H_(2)O`
`CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O," "CO+(1)/(2)O_(2) rarr CO_(2)`
80 mL ( `CH_(4) ` and `CO)` will produce `80 mL CO_(2),C_(3)H_(8)` will produce `=3xx20=60mL`
Total `CO_(2)` produce `=80+60=140`