Question

If a gaseous mixture is made by combining 1.16 g Ar and 2.13 gKr in an evacuated 2.50 L container at 25.0 C, what are the partial pressures of each gas, PAr and PKr, and what is the total pressure exerted by the gaseous mixture?

Answer 1

Ar partial pressure = 28.4 atm

Kr partial pressure = 0.49 atm

total pressure = 28.9 atm

Step-by-step explanation:

1. convert g to moles by dividing by the molar mass of each element

Ar: 116/40=2.9 mol

Kr: 2.13/40=0.05 mol

2. you will need to do this 2 times, one for each substance. plug the number of moles into the ideal gas law PV=nRT to find the partial pressures

Ar: P(2.50L)=(2.9 mol)(0.08206)(25+273 K) —> 28.4 atm Ar

Kr: P(2.50L) = (0.05 mol)(0.08206)(25+273 K) —> 0.49 atm Kr

3. Add the partial pressures to find the total pressure

28.4 + 0.49 = 28.9 atm