chem homework please help!

Question

If a gaseous mixture is made by combining  and  in an evacuated 2.50 L container at  what are the partial pressures of each gas,  and  and what is the total pressure,  exerted by the gaseous mixture?

Answer 1

1. The first step is to convert the number of grams to moles.
2. With the total number of moles known, you can find the total pressure using the ideal gas law.
3. Finally the partial pressures are in the same fraction as the mole ratio. We can use the mole fraction to calculate the partial pressures.

1.39 g Ar * 1 mole/ 39.9 grams = 0.0348 moles Ar

2.45 g Kr *1 mole/ 83.8 grams = 0.0292 moles Kr total moles = 0.0641 moles.

PV = nRT or P = nRT/V = 0.0641 * 0.08206 * 298/2.5 = 0.627 ATM.

partial pressure = mole fraction of gas * total pressure.

partial pressure of Ar = 0.0348/0.0641 * 0.627 = 0.341 Atm

partial pressure of Kr = 0.0292/0.0641 * 0.627 = 0.286 Atm

Answer 2

There are a couple of ways to go about this. The following is one way.

Use the ideal gas law to find the pressure of each gas. Add them up to get the total pressure.

For Ar we have...

1.39 g Ar x 1 mol/40 g= 0.0348 moles

PV = nRT

P = nRT/V = (0.0348)(0.0821)(298K)/2.50 = 0.341 atm

For Kr, we have...

2.45 g Kr x 1 mol/83.8 g = 0.0292 moles

PV = nRT

P = nRT/V = (0.0292)(0.0821)(298)/2.50 = 0.286 atm

Total pressure = 0.341 atm + 0.286 atm = 0.627 atm