When 80 grams of aluminum is reacted with excess chlorine gas, how many formula units of alcl3 are produced?

Al reacts with Cl2 and produces AlCl3. The balanced equation for the reaction is

2Al + 3Cl₂ → 2AlCl₃

Moles = mass / molar mass

Molar mass of Al = 27 g/mol

Mass of Al = 80 g

Hence moles of Al = 80 g / 27 g/mol = 2.96 mol

Stoichiometric ratio between Al and AlCl₃ is 1 : 1

Hence, moles of AlCl₃ = moles of Al = 2.96 mol

Hence, produced AlCl₃ is 2.96 mol 

Amount of AlCl₃ in formula units = moles of AlCl₃ x Avogadro's Constant

                                                        = 2.96 mol x 6.022 × 10²³ mol⁻¹

                                                        = 1.783 x 10²⁴

Hence, 1.783 x 10²⁴ of formula units of AlCl₃ are produced.