A Gaseous mixture contains 4.08g of oxygen and 2.78g of Nitrogen
a)What pressure is exerted by the mixture inside a 100.0ml metal container at 150C.
b) what pressure is contributed by the oxygen?
(please show steps)
Convert g O2 and g N2 to mols. mols = grams/molar mass
Then add mols and use PV = nRT to final total P. Remember T must be in kelvin.
To find P of individual gases, use n for that gas and PV = nRT.
To find the pressure exerted by the gaseous mixture inside the metal container, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the container
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
Step 1: Convert the given temperature from Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 150 + 273.15
T(K) = 423.15 K
Step 2: Calculate the total moles of gas in the mixture using the given masses and molar masses of oxygen (O2) and nitrogen (N2).
Molar mass of oxygen (O2) = 32 g/mol
Molar mass of nitrogen (N2) = 28 g/mol
Moles of oxygen (O2) = mass of oxygen / molar mass of oxygen (O2)
Moles of oxygen (O2) = 4.08 g / 32 g/mol
Moles of oxygen (O2) = 0.1275 mol
Moles of nitrogen (N2) = mass of nitrogen / molar mass of nitrogen (N2)
Moles of nitrogen (N2) = 2.78 g / 28 g/mol
Moles of nitrogen (N2) = 0.0993 mol
Total moles of gas = Moles of oxygen (O2) + Moles of nitrogen (N2)
Total moles of gas = 0.1275 mol + 0.0993 mol
Total moles of gas = 0.2268 mol
Step 3: Convert the volume of the container from milliliters to liters.
V(L) = V(mL) / 1000
V(L) = 100.0 mL / 1000
V(L) = 0.1000 L
Step 4: Substitute the values into the Ideal Gas Law equation and solve for pressure (P).
PV = nRT
P * 0.1000 L = 0.2268 mol * (0.0821 L·atm / K·mol) * 423.15 K
P * 0.1000 L = 9.4444 L·atm
P = 9.4444 L·atm / 0.1000 L
P ≈ 94.44 atm
Therefore, the pressure exerted by the gaseous mixture inside the metal container at 150°C is approximately 94.44 atm.
To find the pressure contributed by oxygen, we can use Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of non-reacting gases is the sum of the partial pressures of each gas.
Step 5: Calculate the moles of oxygen (O2) relative to the total moles of gas.
Moles of oxygen (O2) / Total moles of gas = (0.1275 mol / 0.2268 mol) * 100%
Moles of oxygen (O2) / Total moles of gas ≈ 56.25%
Step 6: Calculate the partial pressure of oxygen (O2) using the total pressure and the mole fraction of oxygen.
Partial pressure of oxygen (O2) = Mole fraction of oxygen (O2) * Total pressure
Partial pressure of oxygen (O2) ≈ 0.5625 * 94.44 atm
Partial pressure of oxygen (O2) ≈ 53.13 atm
Therefore, the pressure contributed by the oxygen in the gaseous mixture is approximately 53.13 atm.