Determine the limiting reactant. 2al(s)+3br2(g)→2albr3(s)

Missing question: For each of the following reactions, 25.0 g of each reactant is present initially.
m(Al) = 25 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 25 g ÷ 27 g/mol.
n(Al) = 0,926 mol.
m(Br₂) = 25 g.
n(Br₂) = 25 g ÷ 160 g/mol.
n(Br₂) = 0,156 mol.
Bromine is the limiting reactant, because it has less amount of substance.