2Al + 3Cl2 --> 2AlCl3 If 12.8 g of aluminum reacts with 39.1g of chlorine, how many g of AlCl3 forms? Also what is the limiting reactant(name only, not mass)

Question

I dont understand the limiting reactant part on the homework

Answer 1

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

Answer 2

Hello, Andres,

The balanced equation tells us we need 2 moles of Al for every 3 moles of Cl₂. So let's see if we have enough, or more than enough, Cl₂ given that we have 12.6 g of Al. Determine the moles of each reactant that we have available:

Moles is obtained by dividing the mass by the molar mass [g/(g/mole) = moles]. The required molar ratio of chlorine to aluminum is (1.5 moles Cl₂)/(mole Al), based on the balanced equation. Multiply the 1.5 times the moles of Al (0.474 moles Al) to determine how many moles of Cl₂ would need to be present to fully react with the Al.

This tells us we would need 0.7111 moles of chlorine to fully react, but we only have 0.551 moles of Cl_2. We'll run out of chlorine once the Al is consumed, so chlorine is the limiting reagent. In fact, we can determine how much aluminum remains after the chlorine is consumed. If all of the chlorine reacts, it would consume:

0.551 moles Cl₂ available x (2 moles Al/3 moles Cl₂) = 0.316 moles Al.

Remaining moles of Al = (0.474 - 0.316) = 0.158 moles Al remaining.

Bob