Calculation of Mass of MnO2 Reduced in the Reaction

The given reaction is as follows:
MnO2 + H2C2O4 + 2H+ → CO2 + H2O + Mn2+

The volume of oxalic acid used in the reaction is 35 ml, and its normality is 0.16N.

To calculate the mass of MnO2 reduced in the reaction, we need to use the formula:

Molarity (M) = Number of moles (n) / Volume (in liters) of solution

The number of moles of oxalic acid can be calculated as follows:

n = M x V

where M is the normality of oxalic acid, and V is the volume of oxalic acid used in the reaction.

n = 0.16 x 35/1000

n = 0.0056 moles

According to the stoichiometry of the reaction, 1 mole of MnO2 reacts with 2 moles of oxalic acid. Therefore, the number of moles of MnO2 can be calculated as follows:

0.0056 moles of H2C2O4 reacts with (0.0056/2) moles of MnO2.

= 0.0028 moles of MnO2

The molar mass of MnO2 is 86.94 g/mol. Therefore, the mass of MnO2 reduced in the reaction can be calculated as follows:

Mass of MnO2 = Number of moles x Molar mass

= 0.0028 x 86.94

= 0.243 g (approx.)

Therefore, the mass of MnO2 reduced in the reaction is 0.243 g.

Explanation of the Reaction

The given reaction is a redox reaction in which MnO2 is reduced to Mn2+ and oxalic acid is oxidized to CO2. The reaction takes place in acidic medium and can be represented as follows:

MnO2 + H2C2O4 + 2H+ → CO2 + H2O + Mn2+

In the reaction, MnO2 acts as an oxidizing agent, and H2C2O4 acts as a reducing agent. The reaction occurs in acidic medium as H+ ions are required to balance the charges of the reactants and products. The balanced chemical equation shows that one mole of MnO2 reacts with two moles of H2C2O4 and produces one mole of CO2, one mole of H2O, and one mole of Mn2+. The reaction is exothermic, and the energy released during the reaction is used to drive the reduction of MnO2 and oxidation of H2C2O4. The reaction is widely used in analytical chemistry to determine the concentration of oxalic acid in the sample.

It's 0.25g.