What minimum mass of $K X_{3} [Fe (CN) X_{6}]$ is needed to convert all the iron in a solution containing 0.30 mol of $Fe X^{2 +}$ into Turnbull's blue, $Fe X_{3} [Fe (CN) X_{6}] X_{2}$ ? (K 39, Fe 56, C 12, N 14)

Answer: C

3 Fe X^{2 +} + 2 [Fe (CN) X_{6}] X^{3 -} Fe X_{3} [Fe (CN) X_{6}] X_{2}

, so moles of ferricyanide = 0.30 x 2/3 = 0.20 mol. Molar mass of

K X_{3} [Fe (CN) X_{6}]

= 3(39)+56+6(26) = 329 g/mol. Mass = 0.20 x 329 = 65.8 g.