Aluminum metal reacts with chlorine gas to form solid aluminum trichloride. what mass of chlorine gas is required to react completely with 163 g of aluminum?

Answer is: mass of chlorine gas is 642.04 grams.
Chemical reaction: 2Al + 3Cl₂ → 2AlCl₃.
m(Al) = 163 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 163 g ÷ 27 g/mol.
n(Al) = 6.04 mol.
From chemical reaction: n(Al) : n(Cl₂) = 2 : 3.
n(Cl₂) = 6.04 mol · 3 ÷ 2.
n(Cl₂) = 9.055 mol.
m(Cl₂) = n(Cl₂) · M(Cl₂).
m(Cl₂) = 9.055 mol · 70.90 g/mol.
m(Cl₂) = 642.04 g.