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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 -A. `10%`B. `20%`C. `90%`D. `80%` |
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Answer» Correct Answer - C 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%` |
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