calculate the total volume of all of the combined gases if they were mixed together into a new container(assume at stp): 1.20•10^25 atoms of?

Molar mass of Cl₂ = 35.45 × 2 = 70.90 g/mol

No. of moles of Cl₂ = (3.7 g) / (70.90 g/mol) = 0.0522 mol
No. of moles of CO₂ = 0.6987 mol
No. of moles of He = (1.20 × 10²⁵) / (6.022 × 10²³) = 19.93 mol

In the new container :
Pressure, P = 1 atm
Absolute temperature, T = 273 K
No. of moles of gases, n = (0.0522 + 0.6987 + 19.93) mol = 20.69 mol
Gas constant, R = 0.0821 atm L / (mol K)

PV = nRT
V = nRT/P

Volume of the container, V = 20.69 × 0.0821 × 273 / 1 L = 464 L


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OR:

Molar mass of Cl₂ = 35.45 × 2 = 70.90 g/mol

No. of moles of Cl₂ = (3.7 g) / (70.90 g/mol) = 0.0522 mol
No. of moles of CO₂ = 0.6987 mol
No. of moles of He = (1.20 × 10²⁵) / (6.022 × 10²³) = 19.93 mol

Total no. of moles of gases, n = (0.0522 + 0.6987 + 19.93) mol = 20.69 mol

1 mole of gas occupies a volume of 22.4 L.
Volume of the container = (22.4 L/mol) × (20.69 mol) = 463 L

calcs.. for YOU to do.
.. (1).. (1.20x10^23 atoms He x 1mol / 6.022x10^23 atoms) = __ mol He
.. (2).. (3.7g Cl x 1 mol Cl / 35.45g Cl) = __ mol Cl
.. (3).. total moles gas = __ mol He + __ mol Cl + __ mol CO2
.. (4) .. total moles gas x (22.41L / mol gas @ STP) = __L

learning points from this problem
.. (a).. gases are independent of one another. We simply sum up the moles of all
.. .. . ...the gases to get total moles of gas
.. (b).. remember 22.41 L/mol... if you forget it.. PV = nRT --> V/n = RT/P
.. .. .. .R = 0.08206 Latm/molK... and STP means T = 273.15.. ...P = 1atm

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1 more point... the problem stated.. 3.7g of Cl.... chlorine gas actually exists as Cl2 in nature.... IF the problem statement meant Cl2... or if you were suppose to know Cl2 not Cl.. then this would be calc (2)
.. (2).. (3.7g Cl2 x 1mol Cl2 / 70.9 g Cl2) = __ mol Cl2

adjust accordingly.