How much chlorine would be required to completely displace bromine from 5.95 g of potassium bromide?

Let’s execute the solution:
1. Let’s write the equation in accordance with the condition of the problem:
m = 5.95 g; X l. -?
2KBr + Cl2 = Br2 + KCl – OBP, obtained bromine, potassium chloride;
2. Calculations according to the formulas of substances:
M (KBr) = 119 g / mol;
Y (KBr) = m / M = 5.95 / 119 = 0.05 mol.
3. Let’s make a proportion according to the process data:
0.05 mol (KBr) – x mol (Cl2);
-2 mol – 1 mol from here, X mol (Cl2) = 0.05 * 1/2 = 0.025 mol.
4. Find the volume of Cl2:
V (Cl2) = 0.025 * 22.4 = 0.56 L.
Answer: for the OVR process, chlorine with a volume of 0.56 liters is required.