Fifteen milliliters of gaseous hydrocarbon (A) was required for complete combustion 357 mL of air (21% oxygen by volume) and gaseous products occupicd 327mL (all volumes being measured at STP)

The molecular formula of the hydrocarbon (A) is:

Options

A. `C_2H_6`
B. `C_2H_4`
C. `C_3H_6`
D. `C_3H_8`

Answer

D

Explanation

Let the formula of hydrocarbons A is `CxHy`
`CxHy (g) + (x + (y)/(4)) O_2(g) to x CO_2(g) + (y)/(2) H_2O(l)`
` mL ( x + (y)/(4)) mL " "xmL-`
`15mL ( x+ (y)/(4)) mL " "15xmL-`
volume of `O_2 = (357 xx 21)/(100) = 75 mL `
volume of `N_2 = 357-75=282mL`
volume of `CO_2` = 327-282 = 45mL
`:. 15x - 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)=`C_3H_8`