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.

Thermohaline circulation and gyres

The correct answer would be b) volume versus temperature, because they are also directly proportional.