A mixture containing 2.81 g each of CH4 (g), C2H4 (g) and C4H10 (g) is contained in a 1.50 L flask at a temperature of 35 oC. What is the partial pressure (in atm) of C4H10 (g)?
To find the partial pressure of C4H10 (g), we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's calculate the number of moles for each gas:
For CH4 (g):
n(CH4) = (mass of CH4) / (molar mass of CH4)
n(CH4) = 2.81 g / 16.04 g/mol
n(CH4) = 0.175 mol
For C2H4 (g):
n(C2H4) = (mass of C2H4) / (molar mass of C2H4)
n(C2H4) = 2.81 g / 28.05 g/mol
n(C2H4) = 0.100 mol
For C4H10 (g):
n(C4H10) = (mass of C4H10) / (molar mass of C4H10)
n(C4H10) = 2.81 g / 58.12 g/mol
n(C4H10) = 0.048 mol
Next, let's convert the temperature to Kelvin:
T(K) = 35 oC + 273.15
T(K) = 308.15 K
Now, we can substitute the values into the ideal gas law equation to find the partial pressure of C4H10 (g):
P(C4H10) = (n(C4H10) * R * T) / V
P(C4H10) = (0.048 mol * 0.0821 L·atm/mol·K * 308.15 K) / 1.50 L
P(C4H10) = 1.000 atm
Therefore, the partial pressure of C4H10 (g) is 1.000 atm.
To find the partial pressure of C4H10, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature in Kelvin
First, we need to calculate the number of moles of C4H10 in the mixture. We can use the molar mass of C4H10 to convert grams to moles:
Molar mass of C4H10 (g/mol) = (4 x Molar mass of C) + (10 x Molar mass of H)
= (4 x 12.01 g/mol) + (10 x 1.01 g/mol)
= 40.04 g/mol + 10.10 g/mol
= 50.14 g/mol
Number of moles of C4H10 = mass of C4H10 / molar mass of C4H10
= 2.81 g / 50.14 g/mol
= 0.056 mol
Now, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
= 35°C + 273.15
= 308.15 K
Plug the values into the ideal gas law equation:
P * V = n * R * T
Solve for P:
P = (n * R * T) / V
P = (0.056 mol * 0.0821 L·atm/K·mol * 308.15 K) / 1.50 L
P ≈ 0.930 atm
Therefore, the partial pressure of C4H10 in the mixture is approximately 0.930 atm.
figure the moles of each, then total moles.
PV=totalmales*RT
solve for pressure p
then
partial pressure=p * moleC4H10/totalmoles