Question

10 ml of hydrocarbon burnt with 80 ml of $O_2$. Collected gases weigh 70 ml. On treatment with KOH solution, 50 ml of gas was obtained at same temperature and pressure. What is the hydrocarbon?

Answer 1

As per standard assumptions water is taken to be in liquid form in combustion reactions

Only CO2 is absorbed in KOH

The remaining gaseous mixture contains unused oxygen and CO2 which amounts to 70 ml. after treating with KOH CO2 will get absorbed .Thus Total volume of CO2 produced = 70 -50 = 20 ml

50 ml Oxygen is left unused. thus used oxygen = 80 - 50 = 30 ml

At constant temperature and pressure volume is directly proportional to moles

thus we have 10 moles of hydrocarbon require 30 moles of oxygen and produce 20 moles of CO2.

OR

we have 1 mole of hydrocarbon requires 3 moles of oxygen and produce 2 moles of CO2

thus the compound should contain 2 atoms of carbon.

It can be either ethane, ethene or ethyne

Now balancing the reaction we get the compound to be C2H4 i.e. ethene