March 26, 2017

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I don't understand how to even start this problem:

A mixture of H2(g), O2(g), and 2 mL of H2O is present in a 0.5 L rigid container at 25 degrees Celsius. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1146 mmHg. Vapor pressure of water at 25 degrees Celsius is 24 mmHg. The mixture is sparked and H2 and O2 react until one reactant is completely missing.
1) Identify the reactant remaining and calculate the number of moles of the reactant.
2) Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 degrees Celsius. The equilibrium pressure of water at 90 degrees Celsius is 526 mmHg.
3) Calculate the number of moles of water present as vapor in the container at 90 degrees Celsius.

Let me get you started.
2H2 + O2 ==> 2H2O

So if we have an equal number of mols H2 and O2, we know all of the hydrogen will react and we will have O2 gas remaining unreacted. Or another way to put it is to start with 1 mol O2 and 1 mol H2. 1 mol O2 will require 2 mols H2 to react completely; we don't have 2 mols H2, only 1 mol which makes hyrogen the limiting reagent. So 1 mol H2 will react with 1/2 mol O2 and that uses all of the H2 and leaves some unreacted O2 remaining.
Now about the pressure.
Poxygen + Phydrogen + Pwater vapor = 1146 mm Hg.
Subtract vapor pressure of water from 1146 to obtain the pressure of hydrogen and oxygen at the beginning.

Check my thinking.

OK I get it.

  • AP Chem - to Dr. Bob - ,


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