Answer to Question #237181 in General Chemistry for Myca

Question

Aluminum chloride, AlCl3, can be made by the reaction of aluminum with chlorine according to the following equation: 2 Al + 3 Cl2 ------> 2 AlCl3

What is the limiting reactant if 20.0 grams of Al and 30.0 grams of Cl2 are used and a chemist carried out the reaction and isolated 29.5 g of AlCl3. What is theoretical yield and percent yield?

Answer 1

Find the molecular masses of all species involved.

Al = 26.98 g/mol

Cl2 = 70.90 g/mol

AlCl3 = 133.33 g/mol

Convert the grams into moles.

moles of Al = g/mm = 20.00 g/26.98 g/mol = 0.74 moles of aluminum on hand.

moles of Cl2 = g/mm = 30.00 g/70.90 g/mol = 0.42 moles of chlorine on hand.

Which is the limiting reagent?

IF we use all aluminum then:

2Al/0.74 = 3Cl2/X

X = 0.28 moles of aluminum are needed.

We don't have 1.11 moles of chlorine. We have 0.42 moles of chlorine. Therefore we

will run out of chlorine first. It is the limiting reactant

Use the limiting reactant to cross the ratio bridge and find the moles of AlCl3 that will

be produced.

3Cl2/0.42 mol = 2AlCl3/x

X = 0.28 moles of AlCl3 are produced

Grams of aluminum chloride are found with g = n * mm = 0.28 mol * 133.33 g/mol = 37.33 g

When 20.0 grams of aluminum and 30.0 grams of chlorine are

reacted according to the above equation, the chlorine is the limiting reactant and the

maximum yield of aluminum chloride is 0.28 moles or 37.33 grams.