Question

A mixture of CH4 and C2H2 occupied a certain volume
at a total pressure equal to 63 torr. The same gas mixture
was burnt to CO, and H2O(l). The CO2(g) alone was
collected in the same volume and at the same
temperature, the pressure was found to be 69 torr.
What was the mole fraction of CH4 in the original gas
mixture ?​

Answer 1

Answer:

Let the partial pressure of CH4 be p thus the partial pressure of C2H2 is 63-p

Total pressure of CO2 after combustion=69mm

therefore

p+2(63-p)=69

p=57mm

partial pressure of C2H2=6mm

percentage of CH4 is

=57/63*100

=90.41%