Gaseous StateHard
Question
15 ml of a gaseous hydrocarbon was required for complete combustion in 357ml of air (21% of oxygen by volume) and the gaseous products occupied 327 ml (all volumes being measured at NTP). What is the formula of the hydrocarbon ?
Options
A.C3H8
B.C4H8
C.C5H10
D.C4H10
Solution
CxHy + O2 → xCO2 + 
+ H2O
15 ml
ml
75 ml
× 15 = 75 x + 
x +
= 5 x + 
= 5
3 +
= 5 15 x + 15x + 282 = 327
y = 8 x = 3
Formula = C3H8
+ H2O 15 ml
ml75 ml
× 15 = 75 x + x +
= 5 x + = 53 +
= 5 15 x + 15x + 282 = 327 y = 8 x = 3
Formula = C3H8
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