# When potassium bromide interacted with chlorine, 16 g of bromine were obtained.

**When potassium bromide interacted with chlorine, 16 g of bromine were obtained. Calculate the mass of potassium bromide taken for the reaction.**

Let’s carry out the solution, compose the reaction equation:

1. 2KBr + Cl2 = Br2 + 2KCl – redox reaction, bromine is released;

2. Let’s make calculations according to the formulas of substances:

M (KBr) = 38.01 + 79.9 = 118.9 g / mol;

M (Br2) = 79.9 * 2 = 159.8 g / mol;

3. Determine the amount of moles of bromine:

Y (Br2) = m / M = 16 / 118.9 = 0.1 mol;

4. Let’s make the proportion:

0.1 mol (Br2) – X mol (KBr);

-1 mol -2 mol from here, X mol (KBr) = 0.1 * 2/1 = 0.2 mol;

5. Let’s calculate the mass of potassium bromide:

m (KBr) = Y * M = 0.2 * 118.9 = 23.78 g.

Answer: To carry out the reaction, it took 23.78 g of potassium bromide.