A 5.0 g sample of a natural gas consisting of methane (CH_(4)) and ethylene (C_(2)H_(4)) was burnt in excess oxygen, yielding 14.5 g of CO_(2) and some H_(2)O as products. Calculate the percentage of ethylene in the sample.

A 5.0 g sample of a natural gas consisting of methane (CH_(4)) and eth

1.	A 5.0 g sample of a natural gas consisting of methane (CH_(4)) and ethylene (C_(2)H_(4)) was burnt in excess oxygen, yielding 14.5 g of CO_(2) and some H_(2)O as products. Calculate the percentage of ethylene in the sample.
Answer» Solution :Suppose ethylene `(C_(2)H_(4))` in the MIXTURE = x g The methane `(CH_(4))` in the mixture = (5 - x) g Equations for complete combustion of `CH_(4)` and `C_(2)H_(4)` are : `C_(2)H(4)+3 O_(2) rarr 2CO_(2)+2H_(2)O` `CH_(4)+2O_(2)rarr CO_(2)+2H_(2)O` `28 g C_(2)H_(4)` PRODUCE `CO_(2)=2xx44 g = 88 g` `therefore x g C_(2)H_(4)` will produce `CO_(2)=(88)/(28)xx x g = (22 x)/(7)g` `16 g CH_(4)` product `CO_(2)=44 g` `(5-x)g CH_(4)` will produce `CO_(2)=(44)/(16)xx(5-x)=(11(5-x))/(4)` `therefore (22x)/(7)+(11(5-x))/(4)=14.5`or `88x +77(5-x)=28xx14.5` or `11 x = 406 - 385=21` or `x=(21)/(11)` `therefore %` of `C_(2)H_(4)=(21//11)/(5)xx100=(21)/(55)xx100=38.18 %` .