Question

35.6 g of Zn reacts with 87.1 g of Br to produce ZnBr2, the only product, with no reactant left over. How much ZnBr2 could be formed by reacting 156 g of Br?

1. 538 g
2. 45.4 g
3. 382 g
4. 220 g
5. 0.0156 g
6. 0.00262 g

Answer 1

Too many unanswered questions in this question. 156 g Br2 and how much Zn? Was the yield 100%.

Answer 2

On second thought if there were no other products the reaction must be 100% so the only question is how much Zn reacts with the 156g Br2.If just 35.6g Zn is provided, the product will be the same as with 87.1g Br2; therefore, 35.6/65.38 = 0.544 mols and that x 225.2 (molar mass ZnBr2) = 122.5 g ZnBr2 and that isn't one of the answers.
So 156g Br2 with whatever Zn is required will be mols Br2 = grams/molar mass = 156/159.2 = 0.980 and that x molar mass ZnBr2 = 0.980 x 225.2 = 220.6g ZnBr2. There is an answer close to that.