Find the volume of chlorine released by the interaction of 17.4 g of manganese (IV) oxide with concentrated hydrochloric acid.

univerkov | February 23, 2021 | education

1. Let’s write the reaction equation:

MnO2 + 4HCl = MnCl2 + Cl2 + 2H2O.

2. Find the amount of reacted manganese oxide:

n (MnO2) = m (MnO2) / M (MnO2) = 17.4 g / 87 g * mol = 0.2 mol.

3. According to the reaction equation, we find the amount, and then the volume of released chlorine (Vm – molar volume, constant equal to 22.4 l / mol):

n (Cl2) = n (MnO2) = 0.2 mol.

V (Cl2) = n (Cl2) * Vm = 0.2 mol * 22.4 L / mol = 4.48 L.

Answer: V (Cl2) = 4.48 liters.

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