0.400 g of a sample containing KClO3 (molar mass 122.5) is reduced by 50.0 mL of 0.400 M Fe^2+ in acid (ClO3- + 6Fe^2+ + 6H+ -> Cl- + 6Fe^3+ + 3H2O). The excess Fe^2+ is titrated with 0.0200 M K2Cr2O7, needing 20.0 mL. The percentage of KClO3 in the sample is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx200

0.400 g of a sample containing $KClO X_{3}$ (molar mass 122.5) is reduced by 50.0 mL of 0.400 M $Fe X^{2 +}$ in acid ( $ClO X_{3} X^{-} + 6 Fe X^{2 +} + 6 H X^{+} Cl X^{-} + 6 Fe X^{3 +} + 3 H X_{2} O$ ). The excess $Fe X^{2 +}$ is titrated with 0.0200 M $K X_{2} Cr X_{2} O X_{7}$ , needing 20.0 mL. The percentage of $KClO X_{3}$ in the sample is:

Total

Fe X^{2 +}

= 20.0 mmol. Excess: mmol dichromate = 0.400, each oxidising 6

Fe X^{2 +}

, so excess = 2.40 mmol.

Fe X^{2 +}

used = 17.6 mmol, so

KClO X_{3}

= 17.6/6 = 2.933 mmol. Mass = 2.933 x 122.5 = 359.3 mg; purity = 359.3/400 x 100 = 89.8%.