80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
A. C _ 2 H _ 2
KOH absorbs CO_2. Volume of CO_2 produced = 224 - 64 = 160 mL Unreacted O_2 = 64 mL, so O_2 consumed = 264 - 64 = 200 mL For hydrocarbon C_xH_y: CO_2 Hydrocarbon = 160 80 = 2 x = 2 O_2 consumed per mole = x + y 4 = 200 80 = 2.5 2 + y 4 = 2.5 y = 2 Hydrocarbon is C_2H_2
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