A solution contains 3.36 g of Fe^3+. It is treated with excess K4[Fe(CN)6] and the whole of the iron(III) is converted to Prussian blue, Fe4[Fe(CN)6]3 (4Fe^3+ + 3[Fe(CN)6]^4- -> Fe4[Fe(CN)6]3). Using Fe = 56, C = 12, N = 14, the mass of Prussian blue formed in grams (to two decimal places) is

JEE Main levelNumerical answerDifficulty 2 of 3qc077

A solution contains $3.36 g$ of $Fe X^{3 +}$ . It is treated with excess $K X_{4} [Fe (CN) X_{6}]$ and the whole of the iron(III) is converted to Prussian blue, $Fe X_{4} [Fe (CN) X_{6}] X_{3}$ ( $4 Fe X^{3 +} + 3 [Fe (CN) X_{6}] X^{4 -} \to Fe X_{4} [Fe (CN) X_{6}] X_{3}$ ). Using Fe $= 56$ , C $= 12$ , N $= 14$ , the mass of Prussian blue formed in grams (to two decimal places) is

Numerical answer type, answer in g, tolerance 0.01.

Moles of

Fe X^{3 +} = 3.36/56 = 0.060

. Four

Fe X^{3 +}

give one formula unit, so moles of Prussian blue

= 0.060/4 = 0.015

. Molar mass

= 4 (56) + 3 [56 + 6 (12 + 14)] = 224 + 636 = 860 g mo l^{- 1}

. Mass

= 0.015 \times 860 = 12.90 g