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:

Fifteen milliliters of gaseous hydrocarbon (A) was required for comple

1.	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:
Answer» `C_2H_6` `C_2H_4` `C_3H_6` `C_3H_8` Solution :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`