The formula of a gaseous hydrocarbon which requires 6 times

Question

The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O_(2) for complete oxidation and produces 4 times its own volume of CO_(2) is C_(x)H_(y). The value of y is _____________.

DivineJEE · Accepted Answer

Combustion reaction :C_xH_y(g) + ( x + y / 4 )O_2(g) -> xCO_2(g) + y / 2H_2O(l)Suppose, volume of C_(x)H_(y) is V and volume of O_(2) is 6 times greater than C_(x)H_(y) = 6Vthen volume of xCO_(2) => V_(x) = 4 Vx = 4Since, V_O_2 = 6 x V_C_xH_yV( x + y / 4 ) = 6V( x + y / 4 ) = 6 ..... (i)Put value of x = 4 in Eq. (i) we get,4 + y / 4 = 6 => y = 8

Explanation

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