Answer:
Al2O3 + 6HCl ==> 2AlCl3 + 3H2O ... balanced equation
moles Al2O3 present = 30.0 g x 1 mol/101.96 g = 0.294 moles
moles HCl present = 30 g HCl x 1 mol/36.5 g = 0.822 moles HCl
HCl is LIMITING as it takes 6 moles HCl for each 1 mol Al2O3 and here that is not enough. It will run out first.
Now, using the limiting reactant, we find the moles and mass of AlCl3 that can be formed.
0.822 moles HCl x 2 moles AlCl3/6 moles HCl = 0.274 moles AlCl3 formed
mass of AlCl3 = 0.274 moles AlCl3 x 133 g/mole = 36.4 g AlCl3 formed