Calculating the Equivalent Weight of MnO2

To calculate the equivalent weight of MnO2 in the given reaction, we need to understand the concept of equivalent weight. The equivalent weight of a substance is the weight of the substance that can combine with or displace one equivalent of hydrogen or any other equivalent weight of another substance.

In the given reaction, 2 moles of MnO2 react with 4 moles of HCl to form 2 moles of MnCl2, water, and oxygen. Therefore, the equivalent weight of MnO2 can be calculated as follows:

1. Calculate the molecular weight of MnO2:
Molecular weight of MnO2 = (1 x atomic weight of Mn) + (2 x atomic weight of O)
= (1 x 55) + (2 x 16)
= 87 g/mol

2. Calculate the number of equivalents of MnO2:
1 equivalent of MnO2 = molecular weight of MnO2 / valency of MnO2
= 87 / 4
= 21.75 g/equivalent

Therefore, the equivalent weight of MnO2 in the given reaction is 21.75 g/equivalent.

Explanation of the Reaction

The given reaction involves the oxidation of HCl by MnO2 in the presence of an acid catalyst. The reaction takes place in two stages:

1. MnO2 is reduced to Mn2+:
MnO2 + 4H+ + 2e- → Mn2+ + 2H2O

2. HCl is oxidized to Cl2:
Cl- → Cl2 + 2e-
2H+ + 2Cl- → Cl2 + H2

Overall reaction:
2MnO2 + 4HCl → 2MnCl2 + 2H2O + O2

In this reaction, MnO2 acts as an oxidizing agent and HCl acts as a reducing agent. The acid catalyst (H+) helps in the reduction of MnO2 and the oxidation of HCl. The reaction produces MnCl2, water, and oxygen gas.