A 22.2L flask contains a mixture of nitrogen gas and oxygen gas at 53°C. The total pressure of the gaseous mixture is 0.623 atm and the mixture is known to contain 0.1105 mol N2. Calculate the partial pressure of oxygen and the moles of oxygen present.
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The ratio of the partial pressures is equal to the ratio of the moles.
Partial pressure of oxygen / Partial pressure of nitrogen = Moles of oxygen / Moles of nitrogen
Let’s determine the total number of moles of gas in the flask.
P * V = n * R * T
0.623 * 22.2 = n * 0.0821 * 326
n = (0.623 * 22.2) ÷ (0.0821 * 326)
Number of moles of oxygen = [(0.623 * 22.2) ÷ (0.0821 * 326)] – 0.1105
The number of moles of oxygen is approximately 0.4062
Partial pressure of oxygen / Partial pressure of nitrogen = 0.4062 / 0.1105 = 3.6805
The total pressure is 0.623 atm.
Let x = partial pressure of oxygen, 0.623 – x = partial pressure of nitrogen
x/ (0.623 – x) = 3.6805
Multiply both sides by (0.623 – x)
x = (0.623 – x) * 3.6805
x = (0.623 * 3.6805) – 3.6805 * x
Add 3.6805 * x to both sides
4.6805 * x = (0.623 * 3.6805)
x = (0.623 * 3.6805) ÷ 4.6805
The partial pressure of oxygen is approximately 0.5atm
Partial pressure of oxygen / Partial pressure of nitrogen = Moles of oxygen / Moles of nitrogen
Let’s determine the total number of moles of gas in the flask.
P * V = n * R * T
0.623 * 22.2 = n * 0.0821 * 326
n = (0.623 * 22.2) ÷ (0.0821 * 326)
Number of moles of oxygen = [(0.623 * 22.2) ÷ (0.0821 * 326)] – 0.1105
The number of moles of oxygen is approximately 0.4062
Partial pressure of oxygen / Partial pressure of nitrogen = 0.4062 / 0.1105 = 3.6805
The total pressure is 0.623 atm.
Let x = partial pressure of oxygen, 0.623 – x = partial pressure of nitrogen
x/ (0.623 – x) = 3.6805
Multiply both sides by (0.623 – x)
x = (0.623 – x) * 3.6805
x = (0.623 * 3.6805) – 3.6805 * x
Add 3.6805 * x to both sides
4.6805 * x = (0.623 * 3.6805)
x = (0.623 * 3.6805) ÷ 4.6805
The partial pressure of oxygen is approximately 0.5atm