50 " mL of " a gaseous hydrocarbon A requried for complete combustion. 357 " mL of " air `(21%` oxygen by volume) and gaseous products occupied 327 mL (all volumes being measured at STP. The molecular formula of the hydrocarbon A is
(a). `C_2H_6`
(b). `C_2H_4`
(c). `C_3H_6`
(d). `C_3H_6`

Explanation

Let the formula of the hydrocarbon A be CxHy.
`C_(x)H_(y)(g)+(x+(y)/(4))O_2(g)toxCO_2(g)+(y)/(2)H_2O(1)`
Volume of `O_2=(357xx21)j/(j10j0)=75mL`
Volume of `N_2=357-75=282ml`
volume of `CO_2=327-282=45mL`
`therefore15x=45`,
`x=3`
`({:(15(x+(y)/(4))=75),(x+(y)/(4)=5):})`
`3+(y)/(4)=5`
On solving we get `y=8`
Formula of A is `C_3H_8`