A gas mixture of `CH_(4)` and `C_(3)H_(6)` undergo complete cracking into `C_(s)` and `H_(2)(g)`. The total mass of `H_(2) (g)` produced is 42 gm. If the total volume of the initial gas mixture at 1 atm and `0 .^(@)C` is 224 litre then mole % of `CH_(4)` in original mixture is -

Options

A. `10%`
B. `20%`
C. `90%`
D. `80%`

Answer

C

Explanation

Mole of mixture `= (224)/(22.4) = 10`
`CH_(4) rarr C (s) + 2H_(2) (g)`
`{:(CH_(4),rarr,C (s),+ 2H_(2) (g),),(x "mole",,,2x "mole",),(C_(3)H_(6),rarr,3C (s),+ 3H_(2) (g),),((10 - x) "mole",,,3 (10 - x) "mole",):}`
`[2x + 3 (10 - x )] = 21`
`2x + 30 - 3x = 21`
`x = 9` mole
`%` mole of `CH_(4) = (9)/(10) xx 100 = 90%`