Finding amount yielded in equations

Question

Find the mass of AlCl3 produced when 10.0g Al2O3 react with 10.0g HCl. Al2O3 + 6HCl > 2AlCl3 + 3H2O

show each step

Answer 1

Al₂O₃ + 6HCl ==> 2AlCl₃ + 3H₂O ... balanced equation

Step 1: Find the limiting reactant. An easy way is to divide mols of each reactant by the corresponding coefficient in the balanced equation.

Al2O3: 10.0 g x 1 mol Al2O3 / 102 g = 0.0980 mols (÷1 -> 0.0980)

HCl: 10.0 g x 1 mol HCl / 36.5 g = 0.274 mols (÷6 -> 0.0457)

Since 0.0457 is less than 0.0987, HCl is LIMITING, and it will determine how much product is formed.

Step 2: Use 0.274 mols HCl to mass AlCl₃ formed by using dimensional analysis:

grams AlCl₃ formed = 0.274 mols HCl x 2 mols AlCl₃ / 6 mols HCl x 133 g AlCl₃ / mol = 12.1 g AlCl₃ (3 sig. figs.)

Answer 2

First find the limiting reagent. The number of moles of Al2O3 is 10.0/102=0.1 mol. HCl is 10.0/36.5=0.27 mol. The coefficient of HCl is 6. 6*0.1 >0.27. Therefore HCl is limiting. You can produce 0.27/3=0.09 mol of AlCl3, which is 0.09*133 = 12 g.