If 30.0 grams of AlC3 is produced in the reaction, how many grams of HCI must have reacted?

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