Question

A manganese-containing ore is digested and chemically treated to isolate manganese as MnO₂(s). Heating the MnO₂ at 1000 °C in air converts it entirely to Mn₃O₄(s).
Suppose a 2.70-g ore sample yields 0.235 g of Mn₃O₄(s) by this process.

Calculate the percent Mn in the ore sample.

Calculate the percent MnO₂ in the ore sample.

Answer 1

1000 0 3MnO2 1000 m₂ y + 0 Manoy = 0.235 g 229 g/mol Mmm,ou = 1:03 x103 mol from 0 Samoa = 3 n Mnzon Dino = 3x Mony Mno, = 3.0g x 10 mol Since Moles of Moo₂ = moles of Mr. Mno₂} Sn=3.09810 moles Mass of Mr = 3.09x 103 molx 55 g/mol Mun = 0-16995 g Min = 0.17 og %. Mn = 0.17 og x 100 man of one-sample 0-170 g xicore 2-70 g 7. Mn = 6.30%. 7. Mno2 - man of Mro xoo Man of sample & mans of MnO2 = ano x Molar Mans of Mnoz = 3.0x10 3x 87 mol Olmol Maną = 0 2 0 2 6 9 9 8 / MPO2 = 0.269 x100 2.7 %. MnO₂ = g -96 so.