Question #3d2dd

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Question #3d2dd

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Answer 1 · 2016-03-23T17:33:22

You get a precipitate of iron(III) hydroxide, $F e {(O H)}_{3} (s)$ $Fe(OH)_3(s)$ .

Explanation:

The net equation is:

$O H^{-} (a q) + F e^{3+} (a q) \to F e {(O H)}_{3} (s)$ $OH^"-"(aq) + Fe^"3+"(aq) -> Fe(OH)_3(s)$

This double displacement reaction is driven by the very low solubility of iron(III) hydroxide ( $5.3 \cdot 10^{-5} g / 100 g of water$ $5.3*10^"-5" "g"//"100 g of water"$ ) at 20 °C).

$3 K O H (a q) + F e {(N O_{3})}_{3} (a q) \to F e {(O H)}_{3} (s) + 3 K N O_{3} (a q)$ $3KOH(aq) + Fe(NO_3)_3(aq) -> Fe(OH)_3(s) + 3KNO_3(aq)$

Potassium ( $K^{+} (a q)$ $K^"+"(aq)$ ) and nitrate ( $N O_{3}^{-} (a q)$ $NO_3^"-"(aq)$ ) are spectator (unchanged) ions.

Filter out the iron(III) hydroxide precipitate and you'll have a solution of pure potassium nitrate, $K N O_{3} (a q)$ $KNO_3(aq)$ .

After evaporation of water, if there isn't any excess of iron(III) or hydroxide, we can eventually crystallise a solid sample of the second product: $K N O_{3}$ $KNO_3$ .

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Question #3d2dd

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