# balanced chemical equation for ? "Ni(NO"_3)_2 +"NaCO"_3"rarr"NaNO"_3 + "NiCO"_3"

## how I balance

Feb 25, 2018

$N i {\left(N {O}_{3}\right)}_{2} + N {a}_{2} C {O}_{3} \rightarrow 2 N a N {O}_{3} + N i C {O}_{3}$

#### Explanation:

I believe in your equation, it should be $N {a}_{2} C {O}_{3}$ instead of $N a C {O}_{3}$ because sodium has an oxidation number of $+ 1$ while carbonate ($C {O}_{3}$) has an oxidation number of $- 2$, so two sodium atoms will bond to carbonate.

Balancing a chemical equation means to have the same kinds and amounts of atoms on each side of the equation. So that means there must be equal amounts of nickel, nitrate ($N {O}_{3}^{- 1}$), sodium, and carbonate (CO_3^(-2))molecules on each side.

A strategy to balancing reactions is to start with one inequality and work from there.

For example, in the unbalanced equation:

$N i {\left(N {O}_{3}\right)}_{2} + N {a}_{2} C {O}_{3} \rightarrow N a N {O}_{3} + N i C {O}_{3}$

there are two nitrate molecules in the reactants, but only one on the products. Double the whole $N a N {O}_{3}$ molecule to make $2 N a N {O}_{3}$. Now the nitrate molecules are balanced.

After you make a change, check to see how it affects the whole reaction. There are now 2 $N a$ atoms on the right side.

However, there are already 2 $N a$ atoms on the left side so you don't need to balance the sodium atoms.

Check the number of nickel atoms. One on the reactants, one on the products. Check.

Check the number of carbonate molecules. One on reactants, one on products. Check.

Now that you checked all the atoms and molecules, the balanced equation is:

$N i {\left(N {O}_{3}\right)}_{2} + N {a}_{2} C {O}_{3} \rightarrow 2 N a N {O}_{3} + N i C {O}_{3}$

Feb 25, 2018

$\text{Ni(NO"_3)_2("aq") + "Na"_2"CO"_3("aq")}$$\rightarrow$$\textcolor{m a \ge n t a}{\text{2")"NaNO"_3("aq") + "NiCO"_3("s}} \downarrow$

#### Explanation:

The formula for sodium carbonate is $\text{Na"_2"CO"_3}$ instead of $\text{NaCO"_3}$. The carbonate $\left(\text{CO"_3^(2-)}\right)$ has a ${2}^{-}$ charge, and the sodium ion $\left(\text{Na"^(+)}\right)$ has a ${1}^{+}$ charge. So two sodium ions are needed so that the compound is neutral.

Balance the following equation.

$\text{Ni(NO"_3)_2("aq") + "Na"_2"CO"_3("aq")}$$\rightarrow$"NaNO"_3("aq") + "NiCO"_3("s")darr

The down arrow means that the solid is a precipitate.

Both anions are polyatomic ions; the nitrate ion $\left(\text{NO"_3^(-)}\right)$, and the carbonate ion $\left(\text{CO"_3^(2-)}\right)$. We can consider each of them as a single unit when balancing.

Lets look at the cations first. We have $\text{1 Ni}$ ion on both sides. However there are $2$ $\text{Na}$ ions on the left-hand side and $1$ $\text{Na}$ ion on the right-hand side. So place a coefficient of $2$ in front of $\text{NaNO"_3}$

$\text{Ni(NO"_3)_2("aq") + "Na"_2"CO"_3("aq")}$$\rightarrow$$\textcolor{m a \ge n t a}{\text{2")"NaNO"_3("aq") + "NiCO"_3("s}} \downarrow$

Is this equation balanced? Let's check how many of each ion are on each side of the equation.

Left-hand side $\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .}$ Right-hand side
$\text{1 Ni"^(2+)}$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .}$$\text{1 Ni"^(2+)}$
$\text{2 NO"_3^(-)}$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots}$$\text{2 NO"_3^(-)}$
$\text{2 Na"^(+)}$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$$\text{2 Na"^(+)}$
${\text{1 CO"_3}}^{2 -}$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$${\text{1 CO"_3}}^{2 -}$

The numbers of each ion on both sides of the equation are the same now, so it is balanced.

I am adding a table of solubility rules so you can see how I knew which compounds were soluble and insoluble.