# How do you balance Al(s) + 3Zn(NO_3)_2(aq) -> Al(NO_3)_3(aq) + Zn(s)?

Feb 28, 2016

The balanced equation is

$2 \text{Al(s)" + "3Zn"("NO"_3)_2("aq") → 2"Al"("NO"_3)_3"(aq)" + 3"Zn(s)}$

One way is to replace the ${\text{NO}}_{3}$ groups with $\text{X}$ and then balance by inspection.

We recognize that groups within parentheses usually behave as a unit.

The unbalanced equation is

$\text{Al(s)" + "3Zn"("NO"_3)_2("aq") → "Al"("NO"_3)_3"(aq)" + "Zn(s)}$

Let's replace the ${\text{NO}}_{3}$ groups by single symbols $\text{X}$.

Then the equation becomes

$\text{Al(s)" + "3ZnX"_2("aq") → "AlX"_3"(aq)" + "Zn(s)}$

Balance Zn

We have 3 $\text{Zn}$ atoms on the left, so we need 3 $\text{Zn}$ on the right. Put a $3$ in front of $\text{Zn}$.

$\text{Al(s)" + "3ZnX"_2("aq") → "AlX"_3"(aq)" + color(red)(3)"Zn(s)}$

Balance $\text{X}$

We have 6 $\text{X}$ on the left, so we need 6 $\text{X}$ on the right. Put a $2$ in front of ${\text{AlX}}_{3}$.

$\text{Al(s)" + "3ZnX"_2("aq") → color(blue)(2)"AlX"_3"(aq)" + color(red)(3)"Zn(s)}$

Balance $\text{Al}$

We have fixed 2 $\text{Al}$ atoms on the right, so we need 2 $\text{Al}$ atoms on the left. Put a $2$ in front of $\text{Al}$

$\textcolor{f u c h s i a}{2} \text{Al(s)" + "3ZnX"_2("aq") → color(blue)(2)"AlX"_3"(aq)" + color(red)(3)"Zn(s)}$

The equation should now be balanced. Let's check.

On the left: $\text{2 Al, 3 Zn, 6 X}$
On the right: $\text{2 Al, 6 X, 3 Zn}$

Now we replace all the $\text{X}$s with ${\text{NO}}_{3}$ and get the balanced equation.

$2 \text{Al(s)" + "3Zn"("NO"_3)_2("aq") → 2"Al"("NO"_3)_3"(aq)" + 3"Zn(s)}$