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
Too many unanswered questions in this question. 156 g Br2 and how much Zn? Was the yield 100%.
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.
To determine how much ZnBr2 could be formed by reacting 156 g of Br, we need to use stoichiometry.
First, we calculate the moles of Zn and Br using their respective molar masses. The molar mass of Zn is 65.38 g/mol, and the molar mass of Br is 79.90 g/mol.
Moles of Zn = mass of Zn / molar mass of Zn
Moles of Zn = 35.6 g / 65.38 g/mol ≈ 0.544 mol
Moles of Br = mass of Br / molar mass of Br
Moles of Br = 87.1 g / 79.90 g/mol ≈ 1.09 mol
Now, let's determine the stoichiometric ratio between Zn and Br in the balanced chemical equation for the reaction:
Zn + Br2 → ZnBr2
From the equation, we can see that the ratio is 1:1. This means that 1 mole of Zn reacts with 1 mole of Br to produce 1 mole of ZnBr2.
Since we have 0.544 moles of Zn, the number of moles of Br required for complete reaction is also 0.544 moles.
To find the mass of Br required for complete reaction with 0.544 moles, we use the equation:
Mass of Br = moles of Br × molar mass of Br
Mass of Br = 0.544 mol × 79.90 g/mol ≈ 43.483 g
Now, we can compare this with the given mass of Br (156 g) to see if there is enough Br to react completely.
Since we have more than enough Br (156 g), the limiting reactant in this case is Br. This means that the moles of ZnBr2 formed will be equal to the moles of Br used in the reaction.
Moles of ZnBr2 = moles of Br = 0.544 mol
Finally, we calculate the mass of ZnBr2:
Mass of ZnBr2 = moles of ZnBr2 × molar mass of ZnBr2
Mass of ZnBr2 = 0.544 mol × (65.38 g/mol + 2 × 79.90 g/mol) ≈ 0.544 mol × 225.18 g/mol ≈ 123 g
Therefore, the correct answer is not among the given options. The mass of ZnBr2 that could be formed from reacting 156 g of Br is approximately 123 g.