Jitender Pal
Last Activity: 10 Years ago
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%
