If 3.2 grams of aluminum reacts with 5.4 grams of chlorine gas to produce solid aluminum chloride

Question

1. What mass of aluminum chloride will be produced?
2. Which reactant is the limiting reactant?
3. Which reactant is the excess reactant?

Answer 1

2 Al + 3 Cl₂ --------------- 2 AlCl₃

3.2 grams Al 1 mol Al

26.98 g of Al

0.12 moles of Al

5,4 grams of Chlorine 1 mol Chlorine

71 grams of Chlorine

0.076 mol Chlorine

Divide the number of moles by its coefficient, whichever gives you a lower value is the limiting reactant, which means Chlorine is the limiting reactant. and Aluminum is the excess reactant

Grams of Aluminum chloride produced is

5.4 g of Chlorine 1 mol Chlorine 2 mol Aluminum chloride 133.34 grams

71 grams Chlorine 3 moles Chlorine 1 mol Aluminum chloride

6.76 grams of Aluminum chloride