# Question #37ef7

Jan 7, 2015

Here's what I got.

#### Explanation:

First, your general chemical reaction looks like this:

$C u S {O}_{4 \left(a q\right)} + N {a}_{2} C {O}_{3 \left(a q\right)} \to C u C {O}_{3 \left(s\right)} \downarrow + N {a}_{2} S {O}_{4 \left(a q\right)}$

Copper(II) sulfate, $C u S {O}_{4}$, and sodium carbonate, $N {a}_{2} C {O}_{3}$, are soluble in aqueous solution, which means that they will dissociate into cations and anions.

The reaction produces copper(II) carbonate, $C u C {O}_{3}$, an insoluble solid that precipitates out of solution, and aqueous sodium sulfate, $N {a}_{2} S {O}_{4}$.

The complete ionic equation is:

$C {u}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -} + 2 N {a}_{\left(a q\right)}^{+} + C {O}_{3 \left(a q\right)}^{2 -} \to C u C {O}_{3 \left(s\right)} \downarrow + 2 N {a}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -}$

The net ionic equation, which you get after eliminating spectator ions, is:

$C {u}_{\left(a q\right)}^{2 +} + \textcolor{red}{\cancel{\textcolor{b l a c k}{S {O}_{4 \left(a q\right)}^{2 -}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 N {a}_{\left(a q\right)}^{+}}}} + C {O}_{3 \left(a q\right)}^{2 -} \to C u C {O}_{3 \left(s\right)} \downarrow + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 N {a}_{\left(a q\right)}^{+}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{S {O}_{4 \left(a q\right)}^{2 -}}}}$

$C {u}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -} \to C u C {O}_{3 \left(s\right)} \downarrow$