Question

A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while the temperature is held at 400K, what is the partial pressure of H2O

Answer 1

`p_(H_2O)=2.00atm`

Step-by-step explanation:

Hello!

In this case, according to the following chemical reaction:

`2H_2+O_2\rightarrow 2H_2O`

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

`n_(H_2)=(3.00atm*1.00L)/(0.08206(atm*L)/(mol*K)*400K)=0.0914molH_2 \\\\n_(O_2)=(1.00atm*1.00L)/(0.08206(atm*L)/(mol*K)*400K)=0.0305molH_2`

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

`n_(H_2O)^(by\ H_2) = 0.0914molH_2*(2molH_2O)/(2molH_2) =0.0914molH_2O\\\\n_(H_2O)^(by\ O_2) = 0.0305molO_2*(2molH_2O)/(1molO_2) =0.0609molH_2O`

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

`p_(H_2O)=(0.0609molH_2O*0.08206(atm*L)/(mol*K)*400K)/(1.00L)\\\\ p_(H_2O)=2.00atm`

Best regards!