# Question 44dbb

Jun 28, 2015

You can balance this double replacement reaction by inspection.

#### Explanation:

The unbalanced chemical equation looks like this

$N {a}_{3} P {O}_{4 \left(a q\right)} + B a {\left(N {O}_{3}\right)}_{2 \left(a q\right)} \to N a N {O}_{\textrm{3 \left(a q\right]}} + B {a}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)}$

The easiest way to balance this equation is to treat the phosphate, $P {O}_{4}^{3 -}$, and nitrate, NO_3""^(-)#, polyatomic ions as a single unit.

So, notice that you have 2 phosphate ions on the products' side, but only one on the reactants' side. Multiply sodium phosphate by 2 to balance the phosphate ions.

$\textcolor{red}{2} N {a}_{3} P {O}_{4 \left(a q\right)} + B a {\left(N {O}_{3}\right)}_{2 \left(a q\right)} \to N a N {O}_{\textrm{3 \left(a q\right]}} + B {a}_{3} {\left(P {O}_{4}\right)}_{\textcolor{red}{2} \left(s\right)}$

Now focus on the nitrate ions. Notice that you have 2 nitrate ions on the reactants' side, but only 1 on the products' side. Multiply sodium nitrate by 2 to balance the nitrate ions.

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

Now focus on the sodium and barium atoms. Notice that you have 6 sodium atoms on the reactants' side, but only 2 on the products' side $\to$ multiply the sodium nitrate by 3 to balance the sodium atoms.

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

Now the nitrate ions are unbalanced. Multiply the barium nitrate by 3 to balance the 6 nitrate ions on the products' side with the those on the reactants' side.

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

As it turns out, the barium atoms are balanced, 3 on the reactants' side and 3 on the products' side.