A mixture of gases contains 64g of methane, 64g of oxygen and 64g of sulphur dioxide.
The pressure of the mixture is 2l0kPa. What is the partial pressure (in kPa) of the methane
expected to be?
mols CH4 = g/molar mass = 64/16 = 4
mols O2 = 64/32 = 2
mols SO3 = 64/80 = 0.8
Total mol = 6.8
mol fraction CH4 = XCH4 = 4/6.8 = 0.588 round to 0.59
pCH4 = XCH4*Ptotal
pCH4 = 0.59*210 = ? in kPa.
Isn't it amazing how gases come together to form the best mixture party? Now let's calculate the partial pressure of methane, the life of the party in this case.
To find the partial pressure, we need to determine the mole fraction of methane.
First, let's find the number of moles for each gas.
The molar mass of methane (CH4) is 16 g/mol, so 64 g of methane corresponds to 64 g / 16 g/mol = 4 moles.
Since we have 4 moles of methane, oxygen, and sulfur dioxide, the total number of moles in the mixture is 4 + 4 + 4 = 12 moles.
Now, we can determine the mole fraction of methane. The mole fraction is calculated by dividing the number of moles of methane by the total number of moles in the mixture.
Mole fraction of methane = 4 moles / 12 moles = 1/3.
Finally, we can calculate the partial pressure of methane by multiplying the mole fraction by the total pressure of the mixture.
Partial pressure of methane = (1/3) x 210 kPa = 70 kPa.
So, the expected partial pressure of methane is 70 kPa. And remember, when methane is present, get ready for some gas-tastic fun!
To find the partial pressure of methane, we need to use the mole concept.
First, we need to determine the number of moles of each gas in the mixture. To do this, we divide the given mass of each gas by its molar mass.
The molar mass of methane (CH4) is 16.04 g/mol. Therefore, the number of moles of methane is:
moles of methane = mass of methane / molar mass of methane = 64 g / 16.04 g/mol = 4 moles
Similarly, for oxygen (O2), the molar mass is 32.00 g/mol, so the number of moles of oxygen is:
moles of oxygen = mass of oxygen / molar mass of oxygen = 64 g / 32.00 g/mol = 2 moles
And for sulphur dioxide (SO2), the molar mass is 64.06 g/mol, so the number of moles of sulphur dioxide is:
moles of sulphur dioxide = mass of sulphur dioxide / molar mass of sulphur dioxide = 64 g / 64.06 g/mol = 1 mole
Now, since the pressure is shared among all the gases in the mixture, we can assume that the total pressure is the sum of the partial pressures of each gas. Since we are looking for the partial pressure of methane, we can express its partial pressure as:
partial pressure of methane = (moles of methane / total moles of all gases) x total pressure
Since the total pressure is given as 210 kPa, and the total moles of all gases is 4 + 2 + 1 = 7 moles, we can substitute these values into the equation:
partial pressure of methane = (4 moles / 7 moles) × 210 kPa = 120 kPa
Therefore, the partial pressure of methane is expected to be 120 kPa.
To determine the partial pressure of methane, we need to use the concept of mole ratio. The mole ratio is the ratio between the number of moles of different substances in a chemical reaction.
First, we need to find the number of moles of each gas in the mixture. To do this, we will use the molar mass of each substance.
The molar mass of methane (CH4) is 16 g/mol. Therefore, the number of moles of methane is:
moles of methane = mass of methane / molar mass of methane
= 64 g / 16 g/mol
= 4 moles of methane
The molar mass of oxygen (O2) is 32 g/mol. Therefore, the number of moles of oxygen is:
moles of oxygen = mass of oxygen / molar mass of oxygen
= 64 g / 32 g/mol
= 2 moles of oxygen
The molar mass of sulfur dioxide (SO2) is 64 g/mol. Therefore, the number of moles of sulfur dioxide is:
moles of sulfur dioxide = mass of sulfur dioxide / molar mass of sulfur dioxide
= 64 g / 64 g/mol
= 1 mole of sulfur dioxide
Now, we can use these mole ratios to determine the partial pressure of methane. The partial pressure of methane can be calculated using the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature (assumed to be constant).
Since the volume is given as 2 L and the total pressure of the mixture is 210 kPa, we can rearrange the equation to solve for the partial pressure of methane:
PV = nRT
Partial pressure of methane = (moles of methane / total moles) * total pressure
= (4 moles / (4 moles + 2 moles + 1 mole)) * 210 kPa
= (4 moles / 7 moles) * 210 kPa
= 120 kPa
Therefore, the partial pressure of methane in the mixture is 120 kPa.