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Won't that be just 1/4 of the original pressure?
...........2H2 + O2 ==> 2H2O
I.......30dm^3 + 30dm^3...0
C.........-30...-15.......0
E..........0......15
You had 30 + 30 in a 60 dm^3 container.
After the reaction, and with the same conditions you have 15 dm^3 in the 60 dm^3 container. So the pressure should be just 1/4 (that's 15/60) of the initial pressure.