A gas mixture containing N2 and O2 was kept inside a 2 L container at a temperature of 23 C and a total pressure of 1 atm. The partial pressure of oxygen was .722 atm. How many grams of nitrogen were present in the gas mixture?
Ptotal = pN2 + pO2
You know pO2, solve for pN2.
Substitute pN2 into PV = nRT and solve for n. Then
n = mols = grams/molar mass.
You know mols and molar mass; solve for grams N2.
To determine the number of grams of nitrogen present in the gas mixture, we need to use the ideal gas law equation: PV = nRT, where:
P is the total pressure of the gas mixture,
V is the volume of the container,
n is the number of moles of gas,
R is the ideal gas constant, and
T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin by adding 273 to the Celsius value:
23 C + 273 = 296 K
Now, let's rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
Substituting the given values into the equation:
P = 1 atm (total pressure)
V = 2 L (volume)
R = 0.0821 atm·L/mol·K (ideal gas constant)
T = 296 K (temperature)
n = (1 atm) * (2 L) / (0.0821 atm·L/mol·K * 296 K)
n = 0.0848 mol
Next, we can determine the number of moles of nitrogen by subtracting the moles of oxygen from the total moles:
moles of nitrogen = total moles - moles of oxygen
From the given data, we know that the partial pressure of oxygen (PO2) is 0.722 atm. By Dalton's law, the partial pressure of nitrogen (PN2) is equal to the total pressure (P) minus the partial pressure of oxygen (PO2):
PN2 = P - PO2
Substituting the known values:
PN2 = 1 atm - 0.722 atm
PN2 = 0.278 atm
Since we have the pressure, volume, and temperature, we can use the ideal gas law equation to determine the moles of nitrogen:
nN2 = PN2 * V / (R * T)
nN2 = 0.278 atm * 2 L / (0.0821 atm·L/mol·K * 296 K)
nN2 = 0.0180 mol
Finally, to find the mass of nitrogen present, we need to multiply the number of moles of nitrogen by its molar mass. The molar mass of nitrogen (N2) is approximately 28 g/mol. Therefore:
mass of nitrogen = nN2 * molar mass of N2
mass of nitrogen = 0.0180 mol * 28 g/mol
mass of nitrogen = 0.504 g
So, there are approximately 0.504 grams of nitrogen present in the gas mixture.