Question

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

`NaOH(aq) + NaNO_2(aq) + Al(s) +H_2O(l) -> NH_3(aq) + NaAlO_2(aq)`

Answer 1

You have described a redox process, for which 1/2 equations are the standard approach.
`2Al +4H_2O(l) + NO_2^(-) rarr 2AlO_2^(-) +NH_4 ^++2H_2O(l)`

Explanation:

Aluminum metal is oxidized to aluminate, `AlO_2^-` (in water this is probably better represented by `Al(OH)_4^-` but I'll stick with the former):

`Al +2H_2O(l) rarr AlO_2^(-) +3e^(-) +4H^+`. `i`

This is balanced with respect to mass and charge.

Nitrite anion is reduced to ammonia: `N(+III) rarr N(-III)`.

`NO_2^(-) +7H^(+) + 6e^(-) rarr NH_3 +2H_2O(l)`. `ii`

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

And thus `2xx(i)+(ii)`:

`2Al(s) +4H_2O(l) + NO_2^(-) rarr 2AlO_2^(-) +NH_4 ^++2H_2O(l)`

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 `Al^(3+)`; nitrite is reduced to ammonia.

Note that I think basic conditions are required, not acidic condtions. And we could resolve this by adding `1xxHO^-` to each side of the equation (i.e. I want to get rid of the ammonium ion, `NH_4^+`).

`2Al(s) +4H_2O(l) + NO_2^(-) +OH^(-) rarr 2AlO_2^(-) +NH_3+3H_2O(l)`
Charge is balanced; mass is balanced, so I'm happy.