During an experiment, you introduce 5.00g of hydrogen gas and 50.00g of oxygen g

Question

During an experiment, you introduce 5.00g of hydrogen gas and 50.00g of oxygen gas into an otherwise empty 9.00 L steel cylinder. The temperature of the cylinder is maintained at 35 degrees Celcius. The hydrogen is ignited with an electric spark and the following reaction occurs: 2H2(g)+O2(g)---> 2H20

What is the partial pressure of water vapor inside the cylinder after the reaction is complete?

So I calculated the total gas pressure to be 6.96 atm if there are 2.48 mol of H20 left after the reaction, given that H2 is the limiting reagent. How would I go about finding the partial pressure of the water? I'm right on the edge but I can't figure it out.

Explanation / Answer

Amount......2H2(g)         +      O2(g)--->              2H2O

Initial.........5g(2.5mol)........50g(1.5628mol).......0

Final..........0mol.............(1.5625-1.25.............2.5mol

                                           =0.3125mol)

After reaction is complete, no of moles of O2 left = 0.3125 mole

                                                  no of moles of H2O produced = 2.5 mol

The partial pressure is the pressure that a particular gas exerts in a gas mixture.

You have to simply calculate the pressure of water vapor in the cylinder. Presence of O2 in the cylinder does not affect the partial pressure of O2 and the total pressure is the sum of their partial pressures. Their pressure values can be calculated individually according to the equation, PV = nRT

So, PH2O= nRT/V

= 2.5 * 0.082 * 308 / 9

=7.01 atm (Ans)

PO2

= 0.3125*0.082*308/9

=0.88 atm

Total pressure , P = 7.01 + 0.88 =7.89 atm

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