Question

amount of oxygen in litre required for complete combustion of 4 gram ethylene

Answer 1

Firsr, understand that C2H4's and CH4's combustion creates carbon dioxide and water

C2H4 +3O2-> 2CO2+2H2O

CH4 + 2O2 -> CO2 + 2H2O

Then, convert the volumes to mole. It is widely assumed that 1 mole of any gas, at standard room temperature and pressure, occupies 24dm^3. Thus, 50cm^3 is expressed as 50/24000 =0.0021 mole of C2H4, therefore, 0.0021*3 mole of oxygen is required to completely combust this amount of Ethene. 0.0021*3=0.0063mol

Now convert the Methane, it is also 0.0021mol. But the mole ratio of Methane to oxygen is different for this reaction, so only take 0.0021*2 instead of *3, that'll give you 0.0042mol. Thus, we know that the total mole of oxygen required to combust these two compounds is 0.0042+0.0063=0.0105mole

0.0105mole, using the previous concept of 1 mole of gas occupying 24dm^3,

1mol->24dm^3 or 24000cm^3

0.0105 mole -> 24000cm^3*0.0105 = 252cm^3

Therefore, 252cm^3 of oxygen is required for the complete combustion to take place

Please note that I rounded off everything to a reasonable degree of accuracy. The purpose is not to show you the answer but the method to do it, cheers!