One very important concept for this question that we have talked about in our previous session is molar ratio, which is how many moles of reactant is needed to produce a certain amount of moles of product, in this case we have a 2:2 molar ratio, this means that 2 moles of Al for 2 moles of AlCl3
With that information in mind, it is logical to assume that if we have 8.70 moles of AlCl3, we should also have 8.70 moles of Al as reactant
Now we can use the molar mass of Al to indentify how much of mass it is 8.70 moles of Al, the molar mass is 27 g/mol
27 g = 1 mol of Al
x grams = 8.70 moles of Al
x = 235 grams of Al are required