A mixture of gases is prepared from 76.9 g of O2 and 16.0 g of H2. After the reaction of O2 and H2 is complete, what is the total pressure of the mixture if its temperature is 178 ºC and its volume is 11.0 L? What are the partial pressures of the gases remaining in the mixture?
2H2 + O2 ==> 2H2O
mols H2 = grams/molar mass = about 8 but you need to calculate that as well as all of the other values that follow.
mols O2 = about 4.8. What is the limiting reagent? I think that's H2 and it will produce about 8 mols H2O.
How much O2 will be used. That's
8mols H2 x (1 mol O2/2 mol H2) = 8 x (1/2) = 4 so you will have 0.8 mols O2 remaining. Summary. We have no H2 left, 0.8 mols O2, and we have formed 8 mols H2O. Add total mols and substitute into PV = nRT and solve for Ptotal.
Then calculate mole fraction O2 and H2O.
Then pO2 = XO2 x Ptotal
pH2O = XH2O x Ptotal.
To find the total pressure of the mixture, we need to consider the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, let's calculate the number of moles of each gas.
Using the molar mass of O2 (32.00 g/mol), we can calculate the number of moles of O2:
n(O2) = mass(O2) / molar mass(O2)
n(O2) = 76.9 g / 32.00 g/mol
n(O2) = 2.41 mol
Similarly, using the molar mass of H2 (2.02 g/mol), we can calculate the number of moles of H2:
n(H2) = mass(H2) / molar mass(H2)
n(H2) = 16.0 g / 2.02 g/mol
n(H2) = 7.92 mol
Now, let's calculate the total number of moles in the mixture:
n(total) = n(O2) + n(H2)
n(total) = 2.41 mol + 7.92 mol
n(total) = 10.33 mol
Next, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(ºC) + 273.15
T(K) = 178 ºC + 273.15
T(K) = 451.15 K
Now, we can substitute the values into the ideal gas law equation:
P(total) * V = n(total) * R * T
Solving for P(total):
P(total) = (n(total) * R * T) / V
P(total) = (10.33 mol * 0.0821 L·atm/(mol·K) * 451.15 K) / 11.0 L
P(total) = 36.87 atm
Therefore, the total pressure of the mixture is 36.87 atm.
To find the partial pressures of the remaining gases, we can apply Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of its components.
In this case, after the reaction of O2 and H2 is complete, the partial pressure of O2 remaining is equal to its initial partial pressure, and the partial pressure of H2 remaining is also equal to its initial partial pressure.
Therefore, the partial pressures of the remaining gases are:
Partial pressure of O2 = P(O2) = 36.87 atm
Partial pressure of H2 = P(H2) = 36.87 atm
So, both gases have a partial pressure of 36.87 atm in the mixture.