Question #3d2dd

Mar 23, 2016

You get a precipitate of iron(III) hydroxide, $F e {\left(O H\right)}_{3} \left(s\right)$.

Explanation:

The net equation is:

$O {H}^{\text{-"(aq) + Fe^"3+}} \left(a q\right) \to F e {\left(O H\right)}_{3} \left(s\right)$

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

$3 K O H \left(a q\right) + F e {\left(N {O}_{3}\right)}_{3} \left(a q\right) \to F e {\left(O H\right)}_{3} \left(s\right) + 3 K N {O}_{3} \left(a q\right)$

Potassium (${K}^{\text{+}} \left(a q\right)$) and nitrate ($N {O}_{3}^{\text{-}} \left(a q\right)$) are spectator (unchanged) ions.

Filter out the iron(III) hydroxide precipitate and you'll have a solution of pure potassium nitrate, $K N {O}_{3} \left(a q\right)$.

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: $K N {O}_{3}$.