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)?

Question

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)?

Answer 1

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.

Answer 2

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.

Answer 3

figure the moles of each, then total moles.

PV=totalmales*RT
solve for pressure p

then
partial pressure=p * moleC4H10/totalmoles