# Question #c20f5

May 5, 2015

The easiest way to balance this chemical equation is to go by ions, not by atoms.

If you write the complete ionic equation for this reaction, you'll get

$3 N {H}_{4 \left(a q\right)}^{+} + P {O}_{4 \left(a q\right)}^{3 -} + C {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} \to C {a}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} + N {H}_{\textrm{4 \left(a q\right]}}^{+} + C {l}_{\left(a q\right)}^{-}$

If you were to eliminate spectator ions, i.e. the ions that are present on both sides of the equation, you'll get the net ionic equation, which looks like this

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

Notice that you need 3 calcium atoms on the products' side, but only have 1 calcium cation of the reactants' side $\to$ multiply the calcium cation by 3.

Likewise, multiply the phosphate anions by 2 to get them to match the number of phosphate ions present on the products' side.

This will get you

$\textcolor{red}{2} P {O}_{4 \left(a q\right)}^{3 -} + \textcolor{b l u e}{3} C {a}_{\left(a q\right)}^{2 +} \to C {a}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)}$

Now take these stoichiometric coefficients and use them in the overall equation

$\textcolor{red}{2} {\left(N {H}_{4}\right)}_{3} P {O}_{\textrm{4 \left(a q\right]}} + \textcolor{b l u e}{3} C a C {l}_{2 \left(a q\right)} \to C {a}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} + N {H}_{4} C {l}_{\left(a q\right)}$

Now you only have to balance the ammonium ions, $N {H}_{4}^{+}$, and the $C l$. Notice that you have 6 of each on the reactants' side, so multiply the ammonium chloride by 6 to balance everything out

$\textcolor{red}{2} {\left(N {H}_{4}\right)}_{3} P {O}_{\textrm{4 \left(a q\right]}} + \textcolor{b l u e}{3} C a C {l}_{2 \left(a q\right)} \to C {a}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} + \textcolor{g r e e n}{6} N {H}_{4} C {l}_{\left(a q\right)}$