# How do you balance Mg(NO_3)_2 + K_3PO_4 -> Mg_3(PO_4)_2 + KNO_3?

Mar 7, 2017

$3 M g {\left(N {O}_{3}\right)}_{2} + 2 {K}_{3} P {O}_{4} \rightarrow M {g}_{3} {\left(P {O}_{4}\right)}_{2} + 6 K N {O}_{3}$

#### Explanation:

Is the given reaction stoichiometrically balanced? For every reactant particle, is there a corresponding product particle. If there is not, then the reaction cannot be accepted as a representation of chemical (i.e. physical) reality.

Mar 10, 2017

$3 M g {\left(N {O}_{3}\right)}_{2}$ + $2 {K}_{3} P {O}_{4}$ $\rightarrow$ $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + )6KNO_3

#### Explanation:

Let's assume that the chemical reaction will progress as written.

Balancing ionic equations can be quite easy AS LONG AS YOU MEMORIZE YOUR CATIONS AND ANIONS. There are a number of ways to do this, so you have to find a way that actually works for you. For now, let's do a tally sheet of all ions involved.

$M g {\left(N {O}_{3}\right)}_{2}$ + ${K}_{3} P {O}_{4}$ $\rightarrow$ $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $K N {O}_{3}$

Left side:
$M {g}^{2 +}$ = 1
$N {O}_{3}^{- 1}$ = 2
${K}^{1 +}$ = 3
$P {O}_{4}^{- 3}$ = 1

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1
${K}^{1 +}$ = 1
$P {O}_{4}^{- 3}$ = 2

Let's balance the most complicated ion first: the $P {O}_{4}^{3 -}$.

Left side:
$M {g}^{2 +}$ = 1
$N {O}_{3}^{- 1}$ = 2
${K}^{1 +}$ = 3
$P {O}_{4}^{- 3}$ = 1 x $\textcolor{red}{2}$ = 2

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1
${K}^{1 +}$ = 1
$P {O}_{4}^{- 3}$ = 2

Since the $P {O}_{4}^{3 -}$ ion is bonded to the ${K}^{1 +}$ ion, you need to multiply it by 2 as well.

Left side:
$M {g}^{2 +}$ = 1
$N {O}_{3}^{- 1}$ = 2
${K}^{1 +}$ = 3 x $\textcolor{red}{2}$ = 6
$P {O}_{4}^{- 3}$ = 1 x $\textcolor{red}{2}$ = 2

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1
${K}^{1 +}$ = 1
$P {O}_{4}^{- 3}$ = 2

$M g {\left(N {O}_{3}\right)}_{2}$ + $\textcolor{red}{2} {K}_{3} P {O}_{4}$ $\rightarrow$ $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $K N {O}_{3}$

Now notice that this move have created an imbalance in your ${K}^{1 +}$ ion with 6 ions on the left as opposed to only one ion on the right. To compensate, we have:

Left side:
$M {g}^{2 +}$ = 1
$N {O}_{3}^{- 1}$ = 2
${K}^{1 +}$ = 3 x 2 = 6
$P {O}_{4}^{- 3}$ = 1 x 2 = 2

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1
${K}^{1 +}$ = 1 x $\textcolor{b l u e}{6}$ = 6
$P {O}_{4}^{- 3}$ = 2

Again, since the ${K}^{1 +}$ ion is bonded to the $N {O}_{3}^{- 1}$, we need to also multiply it by 6.

Left side:
$M {g}^{2 +}$ = 1
$N {O}_{3}^{- 1}$ = 2
${K}^{1 +}$ = 3 x 2 = 6
$P {O}_{4}^{- 3}$ = 1 x 2 = 2

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1 x $\textcolor{b l u e}{6}$ = 6
${K}^{1 +}$ = 1 x $\textcolor{b l u e}{6}$ = 6
$P {O}_{4}^{- 3}$ = 2

$M g {\left(N {O}_{3}\right)}_{2}$ + $\textcolor{red}{2} {K}_{3} P {O}_{4}$ $\rightarrow$ $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $\textcolor{b l u e}{6} K N {O}_{3}$

Now your $N {O}_{3}^{- 1}$ ion is unbalanced with only 2 ions on the left as opposed to the 6 on the right. Thus,

Left side:
$M {g}^{2 +}$ = 1
$N {O}_{3}^{- 1}$ = 2 x $\textcolor{g r e e n}{3}$ = 6
${K}^{1 +}$ = 3 x 2 = 6
$P {O}_{4}^{- 3}$ = 1 x 2 = 2

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1 x 6 = 6
${K}^{1 +}$ = 1 x 6 = 6
$P {O}_{4}^{- 3}$ = 2

Same as above the $N {O}_{3}^{- 1}$ ion is bonded with your $M {g}^{2 +}$. Therefore,

Left side:
$M {g}^{2 +}$ = 1 x $\textcolor{g r e e n}{3}$ ==3
$N {O}_{3}^{- 1}$ = 2 x $\textcolor{g r e e n}{3}$ = 6
${K}^{1 +}$ = 3 x 2 = 6
$P {O}_{4}^{- 3}$ = 1 x 2 = 2

Right side:

$M {g}^{2 +}$ = 3
$N {O}_{3}^{- 1}$ = 1 x 6 = 6
${K}^{1 +}$ = 1 x 6 = 6
$P {O}_{4}^{- 3}$ = 2

$\textcolor{g r e e n}{3} M g {\left(N {O}_{3}\right)}_{2}$ + $\textcolor{red}{2} {K}_{3} P {O}_{4}$ $\rightarrow$ $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $\textcolor{b l u e}{6} K N {O}_{3}$ (balanced)