Mixture Composition: PV=nRT Problem,1 atm. 0.4 atm.

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The discussion centers on calculating the mole percent composition of a gas mixture of oxygen and hydrogen before and after ignition, where the total pressure of the mixture is 1 atm and the remaining hydrogen pressure is 0.4 atm. The user employs the ideal gas law, PV=nRT, to relate pressure and moles, noting that the volume, gas constant, and temperature remain constant. The conclusion drawn is that the original mixture consisted of 60% oxygen and 40% hydrogen by mole percent, as 0.6 atm of the mixture was consumed during the reaction.

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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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This is not advanced physics, but basic chemistry.

In what molar ratio do hydrogen and oxygen react?

There ir 0.4 atm of hydrogen left. 0.6 atm of mixture was consumed.
 

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