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Igniting MnO2 converts it quantitatively to Mn3O4. A sample of pyrolusite is of the following composition : MnO2 80%, SiO2 and other inert constituents 15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample?[O = 16, Mn = 54.9]
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%
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