Excess aqueous Fe^3+ is treated with K4[Fe(CN)6] to give insoluble Prussian blue, 4Fe^3+ + 3[Fe(CN)6]^4- -> Fe4[Fe(CN)6]3. The moles of K4[Fe(CN)6] required to precipitate 1.0 mol of Fe^3+ completely are

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc200

Excess aqueous $Fe X^{3 +}$ is treated with $K X_{4} [Fe (CN) X_{6}]$ to give insoluble Prussian blue, $4 Fe X^{3 +} + 3 [Fe (CN) X_{6}] X^{4 -} Fe X_{4} [Fe (CN) X_{6}] X_{3}$ . The moles of $K X_{4} [Fe (CN) X_{6}]$ required to precipitate 1.0 mol of $Fe X^{3 +}$ completely are

By the equation, 3 mol of hexacyanoferrate(II) are used per 4 mol of

Fe X^{3 +}

, so

3/4 = 0.75

mol per mol of

Fe X^{3 +}

. Charge check:

4 (+ 3) + 3 (- 4) = 0

. The value 1.33 is the inverted ratio.