0.600 g of pyrolusite ore is heated with 50.0 mL of 0.100 M oxalic acid and dilute H2SO4: MnO2 + H2C2O4 + 2H+ -> Mn^2+ + 2CO2 + 2H2O. The unreacted oxalic acid needs 16.0 mL of 0.0200 M KMnO4. Taking MnO2=87, the percentage of MnO2 in the ore is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq212

$0.600 g$ of pyrolusite ore is heated with $50.0 mL$ of $0.100 M$ oxalic acid and dilute $H X_{2} SO X_{4}$ : $MnO X_{2} + H X_{2} C X_{2} O X_{4} + 2 H X^{+} Mn X^{2 +} + 2 CO X_{2} + 2 H X_{2} O$ . The unreacted oxalic acid needs $16.0 mL$ of $0.0200 M$ $KMnO X_{4}$ . Taking $MnO X_{2} = 87$ , the percentage of $MnO X_{2}$ in the ore is:

Oxalic acid taken

= 5.00

mmol. Left over:

16.0 \times 0.0200 \times 5/2 = 0.800

mmol. Oxalic acid used by

MnO X_{2}

= 4.20

mmol, which equals mmol of

MnO X_{2}

(1:1). Mass

= 4.20 \times 87 = 365.4

mg; percentage

= 365.4/600 \times 100 = 60.9%