Igniting MnO2 in air converts it quantitatively to Mn2O4

Solution:

3 MnO₂ → Mn₃O₄

3 (54.9 + 32) (3 * 54.9 + 64)

= 260.7 g = 228.7 g

Let the amount of pyrolusite ignited = 100.00 g

∴ Wt. of MnO2 = 80 g (80% of 100 g = 80 g)

Wt. of SiO2­ and other inert substances = 15 g

Wt. of water 100 – (80 + 15) = 5 g

According to equation,

260.7 g of MnO2 gives = 228.7 g of Mn3O4

∴ 80 g of MnO2 gives = 228.7/260.7 * 80 = 70.2 g of Mn3O4

NOTE :

During ignition, H2O present in pyrolusite is removed while silica and other inert substances remain as such.

∴ Total wt. of the residue = 70.2 + 15 = 85.2 g

Calculation of % of Mn in ignited Mn3O4

3 Mn = Mn3O4

3 * 54.9 = 164.7 g 3 * 54.9 + 64 = 228.7g

Since, 228.7 g of Mn3O4 contains 164.7 g of Mn

70.2 g of Mn3O4 contains = 164.7/228.7 * 70.2 = 50.55 g of Mn

Weight of resdue = 85.2 g

Hence, percentage of Mn is the ignited sample

= 50.55/85.2 * 100 = 59.33%