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 :
- C2H2
- C2H4
- C4H10
- C2H6
Answer
(1) C2H2
Solution
CxHy + (x + y4 )O2 xCO2 + y2H2O(l)
Gas left after KOH treatment = unused O2 = 64 mL
Volume of CO2 (absorbed by KOH) = 224 − 64 = 160 mL
80x = 160 x = 2
O2 used = 264 − 64 = 200 mL
80 (x + y4 ) = 200 x + y4 = 2.5 y = 2
Hydrocarbon = C2H2
Answer: (1)