# 2Al+Zn(NO3)2=2Al(NO3)3+Zn net ionic equation?

Nov 9, 2015

$2 A {l}^{\circ} \left(s\right)$ + $3 Z {n}^{\text{2+}}$ = $2 A {l}^{\text{3+}}$ + $3 Z {n}^{\circ} \left(s\right)$ (net ionic equation)

#### Explanation:

The equation given above is wrongly balanced, so I have to correct it first.

$A {l}^{\circ}$ + $Z n {\left(N {O}_{3}\right)}_{2}$ = $A l {\left(N {O}_{3}\right)}_{3}$ + $Z {n}^{\circ}$ (unbalanced)

Tallying the atoms based on subscripts,

left side: $A l$ = 1; $Z n$ = 1; ${\left(N {O}_{3}\right)}_{3}$ = 2
right side: $A l$ = 1; $Z n$ = 1; ${\left(N {O}_{3}\right)}_{3}$ = 3

Notice that I am considering the $N {O}_{3}^{-}$ ion as one "atom" in order to avoid confusing myself.

Balancing the equations we have:

$\textcolor{b l u e}{2} A {l}^{\circ} \left(s\right)$ + $\textcolor{red}{3} Z n {\left(N {O}_{3}\right)}_{2}$ = $\textcolor{g r e e n}{2} A l {\left(N {O}_{3}\right)}_{3}$ + $\textcolor{m a \ge n t a}{3} Z {n}^{\circ} \left(s\right)$ (balanced)

left side: $A l$ = (1 x $\textcolor{b l u e}{2}$) = 2; $Z n$ = (1 x $\textcolor{red}{3}$) = 3; ${\left(N {O}_{3}\right)}_{3}$ = (2 x $\textcolor{red}{3}$) = 6
right side: $A l$ = (1x $\textcolor{g r e e n}{2}$) = 2; $Z n$ = (1 x $\textcolor{m a \ge n t a}{3}$) = 3; ${\left(N {O}_{3}\right)}_{3}$ = (3 x $\textcolor{g r e e n}{2}$) = 6

Now that we have the correct balance equation, we must rewrite the equation showing all the ions involved in the reaction.

$2 A {l}^{\circ} \left(s\right)$ + $3 Z {n}^{\text{2+}}$ + $6 N {O}_{3}^{-}$ = $2 A {l}^{\text{3+}}$ + $6 N {O}_{3}^{-}$ + $3 Z {n}^{\circ} \left(s\right)$

Notice that for the $N {O}_{3}^{-}$, I multiply the subscript with the substance's coefficient. Now let us get rid of your spectator ions (ions that are present in both sides of the equation).

$2 A {l}^{\circ} \left(s\right)$ + $3 Z {n}^{\text{2+}}$ + $\cancel{6 N {O}_{3}^{-}}$ = $2 A {l}^{\text{3+}}$ + $\cancel{6 N {O}_{3}^{-}}$ + $3 Z {n}^{\circ} \left(s\right)$

Now let's show the electrons per half-reaction:

$2 A {l}^{\circ} \left(s\right)$ = $2 A {l}^{\text{3+}}$ + $6 {e}^{-}$

$3 Z {n}^{\text{2+}}$ + $6 {e}^{-}$ = $3 Z {n}^{\circ} \left(s\right)$

Thus, the net ionic equation is

$2 A {l}^{\circ} \left(s\right)$ + $3 Z {n}^{\text{2+}}$ = $2 A {l}^{\text{3+}}$ + $3 Z {n}^{\circ} \left(s\right)$