`10 mL` of a mixture of `CH_(4)` and `C_(3)H_(8)` requires `41 mL` of oxygen for complete combustion. What is the volume of `CH_(4)` and `C_(3)H_(8)` in. the mixture.

`10 mL` of a mixture of `CH_(4)` and `C_(3)H_(8)` requires `41 mL` of

1.	`10 mL` of a mixture of `CH_(4)` and `C_(3)H_(8)` requires `41 mL` of oxygen for complete combustion. What is the volume of `CH_(4)` and `C_(3)H_(8)` in. the mixture.
Answer» Suppose the volume `CH_(4)` in `(CH_(4)+C_(3)H_(8))` mix `=x c.c.` `=` Volume of `C_(3)H_(8)` will be `=10-x c.c.` For `CH_(4) CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O` `because 1` vol. of `CH_(4)` requires `2` vol. of `O_(2)` for complete combustion `therefore x c.c.` of `C_(4), 2x c.c.` of `O_(2)` For `C_(3)H_(8) C_(3)H_(8)+5O_(2)rarr3CO_(2)+4H_(2)O` `because 1` volume of `C_(3)H_(8)` requires `5 ml` of `O_(2)` for complete combustion `therefore (10-x)c.c.` of `C_(3)H_(8)` requires `5(10-x)c.c.` of `O_(2)` Total volume of `O_(2)=2x+5(10-x)` it is equivalent to `41` (aC Cording to question) `because 2x+5(10-x)=41` `therefore x=3 c.c.` Volume of `CH_(4)` is `3c.c.` and volume of `C_(3)H_(8)` is `7 c.c.`

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`10 mL` of a mixture of `CH_(4)` and `C_(3)H_(8)` requires `41 mL` of oxygen for complete combustion. What is the volume of `CH_(4)` and `C_(3)H_(8)` in. the mixture.

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