Consider the following reaction under basic conditions: NO-2 (aq) + Al (s) ⟶NH3 (g) + AlO-2 (aq) How many hydroxide ions will appear in the balanced equation? Enter your answer as a whole number without any decimal places.

1

Explanation:

I find that the easiest way to balance an equation in basic solution is first to balance as if it were in acid and then convert to base.

Your skeleton equation is

NO₂⁻ + Al ⟶ NH₃ + AlO₂⁻

1. Separate the reaction into two half-reactions

NO₂⁻ ⟶ NH₃

Al ⟶ AlO₂⁻

2. Balance all atoms other than O and H

Done

3. Balance O

NO₂⁻ ⟶ NH₃ + 2H₂O

Al + 2H₂O ⟶ AlO₂⁻

4. Balance H

NO₂⁻ + 7H⁺ ⟶ NH₃ + 2H₂O

Al + 2H₂O ⟶ AlO₂⁻ + 4H⁺

5. Balance charge

NO₂⁻ + 7H⁺ + 6e⁻ ⟶ NH₃ + 2H₂O

Al + 2H₂O ⟶ AlO₂⁻ + 4H⁺ + 3e⁻

6. Equalize electrons transferred

1*[NO₂⁻ + 7H⁺ + 6e⁻ ⟶ NH₃ + 2H₂O]

2*[Al + 2H₂O ⟶ AlO₂⁻ + 4H⁺ + 3e⁻]

7. Add the two half-reactions

1*[NO₂⁻ + 7H⁺ + 6e⁻ ⟶ NH₃ + 2H₂O]

2*[Al + 2H₂O ⟶ AlO₂⁻ + 4H⁺ + 3e⁻]

NO₂⁻ + 2Al + 2H₂O ⟶ NH₃ + 2AlO₂⁻ + H⁺

8. Convert to base

NO₂⁻ + 2Al + 2H₂O ⟶ NH₃ + 2AlO₂⁻ + H⁺

H⁺ + OH⁻ ⟶ H₂O

NO₂⁻ + 2Al + H₂O + OH⁻ ⟶ NH₃ + 2AlO₂⁻

The coefficient of OH⁻ is 1.