# How would you balance the chemical equation by adding coefficients as needed?

## $N a O H \left(a q\right) + N a N {O}_{2} \left(a q\right) + A l \left(s\right) + {H}_{2} O \left(l\right) \to N {H}_{3} \left(a q\right) + N a A l {O}_{2} \left(a q\right)$

Sep 8, 2016

You have described a redox process, for which 1/2 equations are the standard approach.
$2 A l + 4 {H}_{2} O \left(l\right) + N {O}_{2}^{-} \rightarrow 2 A l {O}_{2}^{-} + N {H}_{4}^{+} + 2 {H}_{2} O \left(l\right)$

#### Explanation:

Aluminum metal is oxidized to aluminate, $A l {O}_{2}^{-}$ (in water this is probably better represented by $A l {\left(O H\right)}_{4}^{-}$ but I'll stick with the former):

$A l + 2 {H}_{2} O \left(l\right) \rightarrow A l {O}_{2}^{-} + 3 {e}^{-} + 4 {H}^{+}$. $i$

This is balanced with respect to mass and charge.

Nitrite anion is reduced to ammonia: $N \left(+ I I I\right) \rightarrow N \left(- I I I\right)$.

$N {O}_{2}^{-} + 7 {H}^{+} + 6 {e}^{-} \rightarrow N {H}_{3} + 2 {H}_{2} O \left(l\right)$. $i i$

Again, this is (I think) balanced with respect to mass and charge, and is therefore reasonable.

And thus $2 \times \left(i\right) + \left(i i\right)$:

$2 A l \left(s\right) + 4 {H}_{2} O \left(l\right) + N {O}_{2}^{-} \rightarrow 2 A l {O}_{2}^{-} + N {H}_{4}^{+} + 2 {H}_{2} O \left(l\right)$

I think this is balanced; but don't trust my arithmetic. If you need help with individual steps, ask for it. All I have done is write 2 half equations. Aluminum is oxidized to $A {l}^{3 +}$; nitrite is reduced to ammonia.

Note that I think basic conditions are required, not acidic condtions. And we could resolve this by adding $1 \times H {O}^{-}$ to each side of the equation (i.e. I want to get rid of the ammonium ion, $N {H}_{4}^{+}$).

$2 A l \left(s\right) + 4 {H}_{2} O \left(l\right) + N {O}_{2}^{-} + O {H}^{-} \rightarrow 2 A l {O}_{2}^{-} + N {H}_{3} + 3 {H}_{2} O \left(l\right)$
Charge is balanced; mass is balanced, so I'm happy.