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.