Calculate the mass of manganese oxide (4) and the volume of a hydrochloric acid solution in the mass

Calculate the mass of manganese oxide (4) and the volume of a hydrochloric acid solution in the mass fraction of HCl 36.5% (p = 1.18 g / ml), which will be required to obtain chlorine capable of displacing iodine weighing 50.8 g from potassium iodide.

Let’s write down the reaction equations:
2KI + Cl2 = 2KCl + I2

MnO2 + 4HCl = Cl2 + MnCl2 + 2H2O

Let’s find the amount of chlorine substance:
n (KI) = m (KI) / Mr (KI) = 50.8 / 166 = 0.306 mol;

n (Cl2) = 0.5 * n (KI) = 0.5 * 0.306 = 0.153 mol;

Let’s find the mass of manganese oxide:
n (MnCl2) = (Cl2) = 0.153 mol;

m (MnO2) = n (MnCl2) * Mr (MnCl2) = 0.153 * 126 = 19.278 g.

Let’s find the volume of hydrochloric acid:
n (HCl) = 4n (Cl2) = 4 * 0.153 = 0.612 mol;

m (HCl) = n (HCl) * Mr (HCl) = 0.612 * 36.5 = 22.338 g;

V (HCl) = m (HCl) / ρ (HCl) = 22.338 / 1.18 = 18.93 ml.

Answer: m (MnO2) = 19.278 g; V (HCl) = 18.93 ml.