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.