Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)?2AlCl3(s)

What is the maximum mass of aluminum chloride that can be formed when reacting 17.0g of aluminum with 22.0g of chlorine?

Answer 1

work out the mass of AlCl3 that you would get if all of each reagent fully reacted. The one that produces the least product is the limiting reagent and the amount of AlCl3 produced by the limiting reagent is the maximum amount of product possible.

mass AlCl3 possible from 17.0 g Al
= 17.0 g Al x (1 mol Al / 26.98 g Al) x (2 mol AlCl3 / 2 mol Al) x (133.33 g AlCl3 / 1 mol)
= 84 g

mass AlCl3 possible from 22.0 g Cl2
= 22.0 g Cl2 x (1 mol Cl2 / 70.9 g Cl2) x (2 mol AlCl3 / 3 mol Cl2) x (133.33 g AlCl3 / 1 mol)
= 27.58 g

The Cl2 provided yields the least product, thus Cl2 is the limiting reagent and 27.58 g is the maximum possible yield of AlCl3.

Answer 2