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.
