Find the volume of chlorine (Cl2) required to obtain 634 grams of AlCl3.

Let’s write the reaction equation for obtaining AlCl3 from chlorine
2Al + 3Cl2 = 2AlCl3
Find the molar masses of AlCl3 and chlorine
M (AlCl3) = 2 * (27 + 35.5 * 3) = 267 g / mol
M (Cl2) = 3 * (35.5 * 2) = 210 g / mol
We get the proportion:
m (Cl2) / M (Cl2) = m (AlCl3) / M (AlCl3), from here we find the mass of chlorine required to obtain 634 grams of AlCl3:
m (Cl2) = m (AlCl3) * M (Cl2) / M (AlCl3), substituting the numerical values, we have:
m (Cl2) = 634 * 210/267 = 499 g
Now let’s find the volume of chlorine:
m (Cl2) / M (Cl2) = V (Cl2) / Vm (Cl2)
V (Cl2) = m (Cl2) * Vm (Cl2) / M (Cl2), substituting the numerical values, we have:
V (Cl2) = 499 * 3 * 22.4 / 210 = 159.5 l
Answer: 159.5 l