For the reaction ?Al+?Cl2 →?AlCl3 , what is the maximum amount of AlCl3 which could be formed from 6 mol of Al and 7.88 mol of Cl2? Answer in units of g.

Answer 1

2Al + 3Cl₂ ==> 2AlCl₃ ... balanced equation

Limiting reactant (divide moles of each by the respective coefficient):

6 mol Al / 2 = 3

7.88 mol Cl₂ / 3 = 2.63

Cl₂ is LIMITING

Maximum amount of AlCl3:

7.88 mol Cl₂ x 2 mol AlCl₃ / 3 mol Cl₂ x 133 g AlCl₃/mol AlCl₃ = answer in grams