Question

A mixture of gases containing 21.0 g of N2, 106.5 g of Cl2, and 12.0 g of He at 14°C is in a 50.0-L container. What is the partial pressure of N2?

Answer 1

The partial pressure of N2 in the mixture, calculated using the Ideal Gas Law, is found to be 0.424 atm.

Step-by-step explanation:

To calculate the partial pressure of N2 in the mixture, we need to use the Ideal Gas Law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant, and T is the temperature in Kelvin.

First, convert the mass of N2 to moles using its molar mass (28.02 g/mol):

n(N2) = 21.0 g / 28.02 g/mol

= 0.749 moles

Next, convert the temperature to Kelvin:

• T(K) = 14°C + 273.15

= 287.15 K

Now, using the Ideal Gas Law:

• P(N2) = (n(N2) × R × T) / V

P(N2) = 0.424 atm

The partial pressure of N2 in the mixture is 0.424 atm.