5.20 g He, 2.41 g F2, and 1.05 g Ar are placed in a 11.6-L container at 47.7 oC. What is the total pressure in the container?

Molar mass of He = 4.00 g/mol
Molar mass of F₂ = 19.00 × 2 = 38.0 g/mol
Molar mass of Ar = 39.95 g/mol

For the gas mixture in the container :
Total number of moles, n = [(5.20/4.00) + (2.41/38.0) + (1.05/39.95)] mol = 1.39 mol
Volume, V = 11.6 L
Absolute temperature, T = (273.2 + 47.7) K = 320.9 K
Gas constant, L = 0.08206 L atm / (mol K)
Total pressure, P = ? atm

PV = nRT
Then, P = nRT/V

Total pressure, P = 1.39 × 0.08206 × 320.9 / 11.6 = 3.16 atm

Find the moles for each element

then find the pressure of each seperately using PV=nRT

T= 47.7 +273.15
n= moles of each separate element
if you use R=0.08206 you will get pressure in atm
if you use R=8.314 you will get pressure in kpa
V=11.6 L

simply add the 3 pressures together to find the total pressure and youre done