Chemistry Stiochemistry

Question

Bromine can be prepared by adding chlorine to an aqueous solution of sodium bromide. How many grams of bromine are formed if 25.g of sodium bromide and 25.0g of chlorine reacted?

Is this the right answer or this?

A) Cl2 + 2NaBr ==> 2NaCl + Br2

moles NaBr present = 25.0 g x 1 mole/103 g = 0.243 moles

moles Cl2 present = 25.0 g x 1 mole/70.9 g = 0.353 moles

Limiting reactant is NaBr based on the mole ratio of 2 moles NaBr per 1 mole Cl2 in the balanced equation.

Moles Br2 formed: 0.243 mol NaBr x 1 mol Br2/2 mol NaBr = 0.122 mol

mass = 0.122 moles Br2 x 160 g/mol = 19.5 g Br2 formed

Or B)

Cl2(g) +2NaBr(aq) ==> Br2(l) +2NaCl(aq)

25.0g of Cl2 x(1mol/70.09g) = 0.35261 mols Cl2 x (1 mol/ 1 mol) = 0.35261 mols of Br2

0.35261 mols Br2 x (159.8g/1 mol) = 56.34708

G of Br2 = 56.3g

Answer 1

I really love the work you've done for this solution. I'd like you to look at your two answers and see which created more. It created more because the equations you're using aren't checking the limitations that the other reactant has. Both approaches are correct, but the Br_2(g) is formed from the limiting reactant, which you stated is NaBr, and further proved by showing that the NaBr has the capacity to make 0.112 mol Br₂, whe the Cl₂ had the capacity to make .353 mol.

So, the correct answer is A. Good job.