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Name the reagent, if any, that you would add to a solution of iron(III) chloride to precipitate (a) iron(III) hydroxide. (b) iron(III) carbonate. (c) iron(III) phosphate.

Short Answer

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Answer: To precipitate the respective compounds from a solution of iron(III) chloride, the following reagents should be added: (a) Sodium hydroxide (NaOH) for iron(III) hydroxide. (b) Sodium carbonate (Na2CO3) for iron(III) carbonate. (c) Sodium phosphate (Na3PO4) for iron(III) phosphate.

Step by step solution

01

Identify the iron(III) cation

In a solution of iron(III) chloride (FeCl3), the iron(III) cation (Fe^3+) is present. We will need to react it with anions that will form the desired compounds.
02

Precipitating iron(III) hydroxide

To precipitate iron(III) hydroxide (Fe(OH)3), we will need to react the iron(III) cation with hydroxide anions (OH^-). To do that, we can add an aqueous solution of a reagent containing hydroxide anions like sodium hydroxide solution (NaOH). The reaction will be: Fe^3+ (aq) + 3OH^− (aq) → Fe(OH)3 (s)
03

Precipitating iron(III) carbonate

To precipitate iron(III) carbonate (Fe2(CO3)3), we will need to react the iron(III) cation with carbonate anions (CO3^2-). We can do that by adding an aqueous solution of a reagent containing carbonate anions like sodium carbonate solution (Na2CO3). The reaction will be: 2Fe^3+ (aq) + 3CO3^2− (aq) → Fe2(CO3)3 (s)
04

Precipitating iron(III) phosphate

To precipitate iron(III) phosphate (FePO4), we will need to react the iron(III) cation with phosphate anions (PO4^3-). We can do that by adding an aqueous solution of a reagent containing phosphate anions like sodium phosphate solution (Na3PO4). The reaction will be: Fe^3+ (aq) + PO4^3− (aq) → FePO4 (s) In conclusion, the reagents to be added to precipitate the respective compounds are: (a) Sodium hydroxide (NaOH) for iron(III) hydroxide. (b) Sodium carbonate (Na2CO3) for iron(III) carbonate. (c) Sodium phosphate (Na3PO4) for iron(III) phosphate.

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Most popular questions from this chapter

Gold metal will dissolve only in aqua regia, a mixture of concentrated hydrochloric acid and concentrated nitric acid in a 3: 1 volume ratio. The products of the reaction between gold and the concentrated acids are \(\mathrm{AuCl}_{4}^{-}(a q), \mathrm{NO}(g),\) and \(\mathrm{H}_{2} \mathrm{O}\). The equation for this reaction where \(\mathrm{HNO}_{3}\) and \(\mathrm{HCl}\) are strong acids is $$ \begin{aligned} \mathrm{Au}(s)+4 \mathrm{Cl}^{-}(a q)+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) & \longrightarrow \\ \mathrm{AuCl}_{4}^{-}(a q)+\mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O} \end{aligned} $$ (a) What stoichiometric ratio of hydrochloric acid to nitric acid should be used? (b) What volumes of \(12 \mathrm{M} \mathrm{HCl}\) and \(16 \mathrm{M} \mathrm{HNO}_{3}\) are required to furnish the \(\mathrm{Cl}^{-}\) and \(\mathrm{NO}_{3}^{-}\) ions to react with \(25.0 \mathrm{~g}\) of gold?

19\. For an acid-base reaction, what is the reacting species, that is, the ion or molecule that appears in the chemical equation, in the following acids? (a) perchloric acid (b) hydriodic acid (c) nitrous acid (d) nitric acid (e) lactic acid, \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\)

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