What is the composition of the original gas mixture in mole percent?

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Riles246
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Physical Chemistry Problem!

The total pressure of a mixture of oxygen and hydrogen is 1 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.4 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole percent?

I started by using the equation P=nRT/V. Since the V, R and T are the same before and after the reaction I solved the equation for P/n = RT/V so the two equations (for before and after the reaction) could be set equal to each other. Then I got stuck! Please Help!
 
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In this question you have a mixture of 2 gases at STP conditions. You remove all of one and some of the other in a ratio of 1:2. You are left with excess gas of only one component that has a certain pressure.

PV=nRT works if you know your volume. You will notice that you weren't given volume. You could use any volume and solve the problem but try using V=22.4 liters. It helps.
 


I don't need to use a volume since the problem states that the volume stays the same. I can set the initial and final states equal to each other by re-arranging the equation to RT/V=P/n. Then the P/n for the initial and final state of the mixtures should equal each other.
 


What do you think 'n' refers to in the initial state? In the final state?