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 O₂ for complete oxidation and produces 4 times its own volume of CO₂ is C_xH_y

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_xH_y is V and volume of O₂ is 6 times greater than C_xH_y = 6V

then volume of xCO₂ V_x = 4 V

x = ...

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