A solution holds 0.0200 mol of iron(III) ions…

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A solution holds 0.0200 mol of iron(III) ions. A chemist adds exactly the amount of potassium hexacyanoferrate(II), $\text{K}_{4}[\text{Fe}(\text{CN})_{6}]$ ( $368\ \text{g/mol}$ ), that is needed to precipitate all the iron as Prussian blue. The precipitate is $\text{Fe}_{4}[\text{Fe}(\text{CN})_{6}]_{3}$ with $M = 860\ \text{g/mol}$ . Find the amount of potassium hexacyanoferrate(II) needed, its mass, the amount of Prussian blue that forms, and its mass.

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The group scheme is: zero $\text{NH}_4^+$ ; I $\text{Pb}^{2+}$ ; II $\text{Cu}^{2+}$ ; III $\text{Fe}^{3+}$ , $\text{Al}^{3+}$ ; IV $\text{Zn}^{2+}$ , $\text{Ni}^{2+}$ , $\text{Co}^{2+}$ , $\text{Mn}^{2+}$ ; V $\text{Ca}^{2+}$ , $\text{Ba}^{2+}$ ; VI $\text{Mg}^{2+}$ . Zinc, aluminium and lead hydroxides dissolve in excess sodium hydroxide. Copper, nickel, zinc and cobalt hydroxides dissolve in excess ammonium hydroxide as ammines. Iron(III) and hexacyanoferrate(II) make Prussian blue: $4\ \text{Fe}^{3+} + 3\ [\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3$ .