Question

A gaseous mixture of C2H4, C2H6 and CH4 having total volume at 150 ml is subjected to combustion in excess of O2.   If the percentage of CH4 in the original mixture is 20, then the volume of CO2 (in mol ) obtained in this process is​

Answer 1

Answer:

The mixture contains 20% CH4

Total volume of compound = 150ml=0.15 L

Then the amount of CH4 present = 20/100 X 0.15

= 0.03 ltrs

Then the balanced equation,

CH4 + 2 O2 ---------> CO2 + 2 H2O

So, ratio

CH4 : CO2 = 1 : 1

0.03 L / CO2 = 1

CO2 = 0.03 X 1 = 0.03 ltrs