how to calculate initial partial pressure?

Question

Hello.
A mixture of C2H2 and excess H2 in a 1.0L vessel exerts a pressure of 52.0 kPa.
After a spark initiates the rxn: C2H2 + 2H2 --> C2H6
and the products cool to the same temperature as that of the initial mixture, the final pressure is 34.0 kPa.
The partial pressure of C2H2 in the mixture of C2H2 and H2 was???
can someone help me???

Answer 1

If x is the initial partial pressure of C2H2 and y the initial partial pressure of H2, then:

x + y = 52 kPa (1)

H2 is in excess so C2H2 is the limiting reagent and reacts fully.

1 mol of C2H2 reacts with 2 mol of H2 and gives 1 mol of C2H6, so
x kPa of C2H2 reacts with 2x kPa of H2 and gives x kPa of C2H6

So, after the reaction comes to end, there are y - 2x kPa of H2 and x kPa of C2H6 in the vessel.

y - 2x + x = 34, or y - x = 34 kPa (2)

Solving the system of (1) and (2) we get:

x = 9 kPa and y = 43 kPa

So the partial pressure of C2H2 in the mixture was 9 kPa.