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If there was a 500 mL container filled with oxygen and hydrogen so it burns completely (2H:1O), how many moles of hydrogen gas would there be if there if pressure is 101.3 kPa and the temperature is 293 K?

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    $\begingroup$ $PV = nRT$ is your friend in this case. $\endgroup$ – ashu Feb 15 '15 at 12:21
  • $\begingroup$ I know this would work if it was pure hydrogen, but what about mixed with oxygen? $\endgroup$ – Collin Feb 15 '15 at 19:07
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    $\begingroup$ Calculate total moles of both gases with the equation and divide those moles between gases in mole ratio (2:1) that you have given in your question. $\endgroup$ – ashu Feb 16 '15 at 11:00
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Using pv = nrt$$1atm\times(0.5L) = (total moles)(.08206\frac{atm\cdot L}{mol\cdot k})(293K) \\ \text{total moles} = .02079mol $$ There are 3 total moles of gas with 2 of those being hydrogen and 1 being oxygen. This means 66% of the total moles is hydrogen and 33% is oxygen. Soooooo 66% times .02079 is the amount of hydrogen moles and 33% times .02079 is the amount of oxygen. TADA!

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