Chem question please help

Question

A 9.65 L container holds a mixture of two gases at 33 °C. The partial pressures of gas A and gas B, respectively, are 0.429 atm and 0.605 atm. If 0.150 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Ideal gas law

PV = nRT

For gas A: n = PV/RT = (0.429)(9.65) / (0.0821)(306)

n = 0.165 moles A

For gas B: n = PV/RT = (0.605)(9.65) / (0.0821)(306)

n = 0.232 moles B

For gas C: n = 0.150 moles C

Total moles of gas = 0.165 + 0.232 + 0.150 = 0.547 moles

P = nRT/V = (0.547)(0.0821)(306) / 9.65

P = 1.43 atm = total pressure