A 0.535 g sample of pyrolusite ore (impure MnO₂) is treated with 1.42 g oxalic acid (H₂C₂O₄∙ 2H₂O) in an acidic medium.Following this reaction,the excess oxalic acid is titrated with 0.1000 M KMnO₄,36.6 mL being required.What is the % MnO₂ in the ore? The reactions are:
H₂C₂O₄ ∙ 2 H₂O(aq) + MnO₂(s) + 2 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 2 CO₂(g)
5 H₂C₂O₄ ∙ 2H₂O(aq) + 2 MnO₄^-(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 18 H₂O(l) + 10 CO₂(g)

Question

Quizplus · Accepted Answer

The correct answer is A because the balanced chemical equation for the reaction between MnO2 and H2C2O4 is:MnO2(s) + 2 H2C2O4(aq) + 2 H+(aq) → Mn2+(aq) + 4 H2O(l) + 2 CO2(g)From this equation, we can see that 1 mole of MnO2 reacts with 2 moles of H2C2O4. Therefore, the number of moles of H2C2O4 used in the reaction is:1.42 g H2C2O4 * (1 mol H2C2O4 / 126.07 g H2C2O4) = 0.01126 mol H2C2O4Since 2 moles of H2C2O4 react with 1 mole of MnO2, the number of moles of MnO2 in the ore is:0.01126 mol H2C2O4 * (1 mol MnO2 / 2 mol H2C2O4) = 0.00563 mol MnO2The mass of MnO2 in the ore is:0.00563 mol MnO2 * (86.94 g MnO2 / 1 mol MnO2) = 0.489 g MnO2The percentage of MnO2 in the ore is:(0.489 g MnO2 / 0.535 g ore) * 100% = 91.4%

A 0535 G Sample of Pyrolusite Ore (Impure MnO2)is Treated with with 1.42

A 0535 G Sample of Pyrolusite Ore (Impure MnO₂)is Treated with with 1.42