Question

A mixture of 0.220 moles rn, 0.350 moles o2 and 0.640 moles ar has a total pressure of 2.95 atm. what is the pressure of o2? a mixture of 0.220 moles , 0.350 moles and 0.640 moles has a total pressure of 2.95 . what is the pressure of ? 0.649 atm 1.17 atm 0.853 atm 0.969 atm 1.03 atm

Answer 1

Answer:

0.85 atm

Explanation:

We start by getting the total number of moles of all gases.

Total moles = 0.220 + 0.350 + 0.640 = 1.21 mol

Then we get the fraction of Oxygen moles and multiply by the total pressure;

The pressure of O2 = 0.350 mol / 1.21 mol x 2.95 atm  

                                = 0.85 atm

Answer 2

Answer:

Partial pressure of O2 = 0.875 atm

Explanation:

Given:

Moles of Radon gas (Rn) =0.220

Moles of O2 gas = 0.350

Moles of argon gas (Ar) = 0.640

Total pressure = 2.95 atm

To determine:

The partial pressure of O2

Explanation:

As per Dalton's Law, the total pressure exerted by a mixture of gases is equal to the sum of their partial pressures.

Here:

$P(total) = P(Rn) + P(O2) + P(Ar)\\\\where \ the\ partial\ pressure \ is \ given\ as:\\\\partial pressure = mole fraction * total pressure\\\\P(O2) = mole fraction (O2) *P(total)\\\\P(O2) = \frac{moles\ O2}{total\ moles}*P(total)\\ \\P(O2) = \frac{0.350}{1.18} * 2.95=0.875 atm$