If you had excess aluminum, how many moles of aluminum chloride could be produced from 21.0 g of chlorine gas, cl2?

When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:

2Al (s) + 3Cl₂ (g) → 2AlCl₃ (s)

Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:

(21.0 g Cl₂ / 70.906 g/mol Cl₂) x 2 mol AlCl₃ / 2 mol Cl₂ = 0.1974 mol AlCl₃

Thus, we have determined that 0.1974 moles of aluminum chloride can be produced from 21.0 g of chlorine gas.