Hello, Teresa,
These are two steps to answering this question. It is helpful if you could be more specific about what part of the question is most confusing. Otherwise, the explanation for solving it gets too long. I'll work through the problem, but need to leave out some ot the details, due to space and time. Ask if you'd like a particular aspect better explained.
Task 1) We need a balanced equation. Without going through the details, I believe the following works:
3Br2 + 2 Al = 2AlBr3.
This tells us we need 3 moles of bromine for every 2 moles aluminum bromide, for a molar ratio of (3/2) Br2/AlBr3
Task 2) Find the moles of AlBr3 in 73.4 grams. Divide the grams by the molar mass of AlBr3 (266.7 g/mole) to get moles AlBr3. I calculate 0.275 moles.
Multiply 0.275 moles AlBr3 by the molar ratio:
(0.275 moles AlBr3)((3/2) Br2/AlBr3) = 0.413 moles Br2
Convert moles to grams by multiplying by the molar mass of Br2.
(0.413 moles Br2)*(159.8 g/mole) = 65.97 grams Br2
I hope this helps,
Bob
