Hi I need help understanding this solution.
10g H2 and 54 g of o2 are contained in a 10 L flask at 200 degrees Celsius.
Upon sparking what is the final pressure of the container.
a equation
2H2 + 02 -->2H20
5 mol 2 mol
we know H20 is in vapor form.
solution shows H2 is excess, O2 is limiting reagent.
I am confused as to why 4 mol of H20 and 1 mol of H2 is used to calculate pressure
P=n(total)RT/V
thanks!
10g H2 and 54 g of o2 are contained in a 10 L flask at 200 degrees Celsius.
Upon sparking what is the final pressure of the container.
a equation
2H2 + 02 -->2H20
5 mol 2 mol
we know H20 is in vapor form.
solution shows H2 is excess, O2 is limiting reagent.
I am confused as to why 4 mol of H20 and 1 mol of H2 is used to calculate pressure
P=n(total)RT/V
thanks!
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2H2(g) + 02(g) -->2H20(g)
10 g of H2 is 5 mol H2. You say 2 mol O2, but give 54 g of O2 that is1.69 mol? lets go with 2 mol O2 as that is easier.
as you say H2 is in excess. From the equation two mol of H2 reacts with one mol O2. In the reaction you start with 2 mol O2, which is the limiting reactant, so that will react with 4 mol H2 to give 4 mol H2O. The final mixture will contain 4 mol of H2O and 1 mol of H2 that is left over.
The total of # mol of gas after the reaction is 5 mol, the volume is 10 L the temp is 473 K (200C), look up R and calculate the pressure.
10 g of H2 is 5 mol H2. You say 2 mol O2, but give 54 g of O2 that is1.69 mol? lets go with 2 mol O2 as that is easier.
as you say H2 is in excess. From the equation two mol of H2 reacts with one mol O2. In the reaction you start with 2 mol O2, which is the limiting reactant, so that will react with 4 mol H2 to give 4 mol H2O. The final mixture will contain 4 mol of H2O and 1 mol of H2 that is left over.
The total of # mol of gas after the reaction is 5 mol, the volume is 10 L the temp is 473 K (200C), look up R and calculate the pressure.
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You used the sum of the moles of the balanced equation. This would be a mole ratio, but because you are given a quantified mass for each species, you need to determine how many moles are in each species:
10gH2 / 2.02gH2 = 4.95 mol H2
54gO2 / 32.00gO2 = 1.6875 mol O2
Note: Remember that hydrogen and oxygen gases are diatomic, therefore their molar mass is twice their atomic weight.
Now, add mols of H2 and mols of O2 to get a total n value:
4.95+1.6875 = 6.6375 TOTAL moles.
Calculate the total pressure by using the Ideal Gas Law:
PV=nRT
=> P=nRT/V
=> P=(6.6375mols)(0.08206L*atm/mol*K)(473K) / 10L *(see the notes at the bottom)
=> P=25.76 atm
*I used 0.08206L*atm/mol*K for the gas constant, assuming you are solving for pressure in terms of atmospheres. If you are solving for pressure in terms of mmHg, you would use 62.36L*mmHg/mol*K for the gas constant value.
Don't forget to use the total number of moles AND to convert deg.C to Kelvin
Where you round can change the answer. I generally try not to worry about significant figures until the end, as I try to get the most precise answer possible.
10gH2 / 2.02gH2 = 4.95 mol H2
54gO2 / 32.00gO2 = 1.6875 mol O2
Note: Remember that hydrogen and oxygen gases are diatomic, therefore their molar mass is twice their atomic weight.
Now, add mols of H2 and mols of O2 to get a total n value:
4.95+1.6875 = 6.6375 TOTAL moles.
Calculate the total pressure by using the Ideal Gas Law:
PV=nRT
=> P=nRT/V
=> P=(6.6375mols)(0.08206L*atm/mol*K)(473K) / 10L *(see the notes at the bottom)
=> P=25.76 atm
*I used 0.08206L*atm/mol*K for the gas constant, assuming you are solving for pressure in terms of atmospheres. If you are solving for pressure in terms of mmHg, you would use 62.36L*mmHg/mol*K for the gas constant value.
Don't forget to use the total number of moles AND to convert deg.C to Kelvin
Where you round can change the answer. I generally try not to worry about significant figures until the end, as I try to get the most precise answer possible.