When liquid bromine is mixed with aluminum metal, a combination reaction occurs, forming aluminum bromide: 2 Al(s) 3Br2(l) --> 2AlBr3(s) . Calculate the mass of AlBr3 produced when 45.0 g of Br2 is added to 30.0 g of Al. The molar mass of AlBr3 is 266.69 g/mol.

Since you are not told which reactant is the limiting reactant, you should find the mass of AlBr₃ starting from both reactants. The correct answer will be the one which is the lower mass.

To find the mass, you should (1) convert from grams to moles of the reactant (via the molar mass from the periodic table), then (2) convert from moles reactant to moles AlBr₃ (via mole-to-mole ratio from equation), and the (3) convert moles to grams AlBr₃ (via molar mass). The final answer should have 3 sig figs because the given values also have 3 sig figs.

2 Al (s) + 3 Br₂ (l) -----> 2 AlBr₃ (s)

Molar Mass (Al): 26.982 g/mol

30.0 g Al 1 mole 2 moles AlBr₃ 266.69 g
-------------- x ---------------- x ----------------------- x ------------------ = 297 g AlBr₃
26.982 g 2 moles Al 1 mole

Molar Mass (Br₂): 159.808 g/mol

45.0 g Br₂ 1 mole 2 moles AlBr₃ 266.69 g
---------------- x ----------------- x ----------------------- x ------------------ = 50.1 g AlBr₃
159.808 g 3 moles Br₂ 1 mole

Since 50.1 g is smaller than 297 g, Br₂ is the limiting reactant and the mass AlBr₃ produced is 50.1 g.