Determine the volume of 2M hydrochloric acid consumed for the interaction with manganese (IV) oxide

| February 10, 2021 | education

Determine the volume of 2M hydrochloric acid consumed for the interaction with manganese (IV) oxide, if it is known that the chlorine released during this process displaced 25.4 g of iodine from the potassium iodide solution.

Data: mI – mass of displaced iodine (mI = 25.4 g); a solution of 2M (2 mol hydrochloric acid in 1 l (1000 ml) of water).

Const: MI – molar mass of iodine (MI ≈ 254 g / mol); MHCl – molar mass of hydrochloric acid (MHCl ≈ 36.5 g / mol); ρHCl – density of hydrochloric acid (ρHCl = 1.19 g / cm3 = 1.19 g / ml).

1) Amount of iodine substance: νI = mI / MI = 25.4 / 254 = 0.1 mol.

2) The level of the second reaction: 2KI (potassium iodide) + Cl2 (chlorine) = 2KCl (potassium chloride) + I2 (iodine).

3) Amount of chlorine substance: νCl / νI = 1/1 and νCl = νI = 0.1 mol.

4) The level of the first reaction: MnO2 (manganese oxide) + 4HCl (hydrochloric acid) = MnCl2 (manganese chloride) + Cl2 ↑ (chlorine) + 2H2O (water).

5) Amount of hydrochloric acid substance: νHCl / νCl = 4/1 and νHCl = 4 * νCl = 4 * 0.1 = 0.4 mol.

6) The volume of hydrochloric acid: VHCl = Vp * νHCl / νp = 1000 * 0.4 / 2 = 200 ml.

Answer: We spent 200 ml of hydrochloric acid.



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