Igniting MnO 2 converts it quantitatively to Mn 3 O 4 . A sample of pyrolusite is of the following composition : MnO 2 80%, SiO 2 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]

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Jitender Pal · Answer

3 MnO 2 &rarr; Mn 3 O 4 3 (54.9 + 32) (3 * 54.9 + 64) = 260.7 g = 228.7 g Let the amount of pyrolusite ignited = 100.00 g &there4; Wt. of MnO 2 = 80 g (80% of 100 g = 80 g) Wt. of SiO 2 &shy; and other inert substances = 15 g Wt. of water 100 &ndash; (80 + 15) = 5 g According to equation, 260.7 g of MnO 2 gives = 228.7 g of Mn 3 O 4 &there4; 80 g of MnO 2 gives = 228.7/260.7 * 80 = 70.2 g of Mn 3 O 4 NOTE : During ignition, H 2 O present in pyrolusite is removed while silica and other inert substances remain as such. &there4; Total wt. of the residue = 70.2 + 15 = 85.2 g Calculation of % of Mn in ignited Mn 3 O 4 3 Mn = Mn 3 O 4 3 * 54.9 = 164.7 g 3 * 54.9 + 64 = 228.7g Since, 228.7 g of Mn 3 O 4 contains 164.7 g of Mn 70.2 g of Mn 3 O 4 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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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]