Find the mass of AlCl3 that is produced when 30.0 grams of Al2O3 react with 30.0 grams of HCl according to the following equation.

Question

Al2O3(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2O(aq)

Answer 1

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.

(note: an easy way to determine limiting reactant is to divide moles of each reactant by the corresponding coefficient in the balanced equation, and the smaller value represents the limiting reactant)

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

Answer 2

Sydney,

Use the molar mass of Al2O3 to find the moles. Then, use the molar ratio of Al2O3 to AlCl3 from your balanced equation to find how many moles of AlCl3 are formed. Do this again, but starting with HCl. You do it both ways to find which one is your limiting reactant. Whichever gives you the least number of moles of AlCl3 is your limiting reactant and this is the maximum amount of AlCl3 that can be made if you are given 30 grams of each.

Once you know how many moles of AlCl3 can be formed, use the molar mass of AlCl3 to determine the grams.

Does this help?