Question

chlorine and nitrogen gases are mixed in a 2.85 l container at 405 k. the total pressure exerted by the gases is 125 kpa, and the partial pressure of chlorine is 38 kpa. how many moles of nitrogen gas are in the mixture? note: 101.325 kpa = 1 atm. 11 mols 7.5 mols 0.032 mols 0.074 mols

Explanation:

Step1: Calculate the partial pressure of nitrogen

According to Dalton's law of partial pressures, \(P_{total}=P_{Cl_2}+P_{N_2}\). So \(P_{N_2}=P_{total}-P_{Cl_2}\).
\(P_{N_2}=(125 - 38)\text{kPa}=87\text{kPa}\). Convert to atm: \(P_{N_2}=\frac{87}{101.325}\text{atm}\approx0.859\text{atm}\)

Step2: Use the ideal gas law \(PV = nRT\) to find the number of moles of \(N_2\)

The ideal gas law is \(n=\frac{PV}{RT}\), where \(R = 0.0821\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\), \(V = 2.85\text{L}\), \(T=405\text{K}\), and \(P = 0.859\text{atm}\)
\(n=\frac{0.859\times2.85}{0.0821\times405}\)

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Answer:

0.074 mols