How many grams of aluminum chloride could be produced from 35.5g of aluminum and 39.0g of chlorine gas?

The balanced chemical equation for the reaction is: 2Al + 3Cl₂ → 2AlCl₃

Calculate the limiting reactant:

• Moles of Al: 35.5g / molar mass of Al
• Moles of Cl₂: 39.0g / molar mass of Cl₂
• Determine which reactant gives the lower amount of AlCl₃ produced.

Once you know the limiting reactant, use stoichiometry to calculate the grams of AlCl₃ produced.