Question

Aluminium reacts with chlorine gas to form aluminium chloride: 2Al + Cl2➡️ 2AlCl3 Determine the limiting reagent if 34 gm of aluminium reacts with 39 gm of chlorine, also find the amount of aluminium chloride formed​

Answer 1

Answer:

The limiting reactant is the one that will run out first based on how many moles are present and based on the stoichiometry of the balanced equation. One easy way to find it is to find moles of each reactant and then divide that number by the coefficient in the balanced equation. The lowest number tells you the limiting reactant. Of course, you MUST have a correctly balanced equation for any of this to work.

2Al + 3Cl2 ==> 2AlCl3

Find limiting reactant:

moles Al present = 12.8 g Al x 1 mol Al/27.0 g = 0.474 moles Al (÷2 ->0.237)

moles Cl2 present = 31.9 g Cl2 x 1 mol/71 g = 0.449 moles Cl2 (÷3 ->0.149) LIMITING REACTANT

Name of limiting reactant is chlorine

mass of AlCl3 formed = 0.449 mol Cl2 x 2 mol AlCl3 / 3 mol Cl2 x 133 g AlCl3/mol = 39.8 g AlCl3