Question

How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction 2Al+3Cl2→2AlCl3?

Answer 1

Answer:

234.9 grams

Explanation:

We rewrite the equation of the reaction as follows :

2Al + 3Cl2 — 2AlCl3

We have 8.70 moles of Aluminum Chloride.

The mole ratio of Aluminum to Aluminum chloride is 2 : 2 = 1 : 1

The moles of Aluminum that are reacting is therefore given as 8.70 moles.

Molar mass of Aluminum is = 27 grams per mole.

The mass of aluminum is therefore :

27 × 8.7 = 234.9 grams