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Topic: Ap?gas law problem  (Read 4009 times)

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Offline malibucat1

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Ap?gas law problem
« on: October 25, 2006, 08:57:30 PM »
A mixture of H2, O2, and 2 mL of H2O is present in a 0.600 liter rigid container at 27degrees C. The number of moles of H2 outnumber the number of moles of O2 by a 3:2 ratio. The total pressure is 634.7mmHg. The equilibrium vapor pressure of pure water at 25 degrees C is 26.7 mmHg. The mixture is sparked, and H2 and O2 react until one reactant is completely consumed.

Identify the reactant remaining and calculate the number of moles of the reactant remaining. And you have to do it this way:

                                                    2H2+       O2    --> 2H2O
Find initial # moles                         ______     _____      _____
Calculate # moles gained or lost      _______   ______   ________
Calculate final # moles                    ______     _____      ______

Offline Yggdrasil

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Re: Ap?gas law problem
« Reply #1 on: October 25, 2006, 09:05:43 PM »
The forum rules say that you have to show that you've tried to answer the question first.  That way, we know where exactly in the problem you need help.

Offline malibucat1

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Re: Ap?gas law problem
« Reply #2 on: October 25, 2006, 09:21:43 PM »
The only part I understand is that O2 is the left over reactant because the equation is says that 2 mol of H2 reacts w/ 1 mol of O2 to make H2o. And since the ratio of h2 to o2 is 3:2, o2 would be left over. But I don't understand what my next step(s) would be

Offline Yggdrasil

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Re: Ap?gas law problem
« Reply #3 on: October 25, 2006, 09:29:19 PM »
Before the reaction occurs, you are given the volume of gas, the temperature, and the pressure.  From this information, you can find the number of moles of gas in the container (remember to account for both the volume of the liquid water, and the partial pressure of the water vapor).  Using this number and the fact that the ratio of H2:O2 = 3:2, you can figure out the initial number of moles of hydrogen and oxygen.

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