Chemical Equation Balance: NaNO3 + NaOH + Zn -> NH3 + Na2ZnO2 + H2O

[Say17]

Homework Statement

NaNO3 + NaOH + Zn -> NH3 + Na2ZnO2 + H2O

Relevant Equations

NaNO3 + NaOH + Zn -> NH3 + Na2ZnO2 + H2O

Hi everyone,

I don't know where to start with NaNO3 + NaOH + Zn -> NH3 + Na2ZnO2 + H2O

How to balance it or where should I start? Any tricks?

Kind regards,
Say

 

[Borek]

What methods of balancing redox reactions do you know?

 

[DaveE]

These videos may be helpful.
https://www.khanacademy.org/science...l-equations/v/chemical-reactions-introduction

Here's my best trick: Study the material, look at some examples that others have solved, make an attempt to solve your problem, check you answers for consistency, then ask for help about the specific parts that you find confusing.

 

[votingmachine]

I tend to think it is a process rather like finding a common denominator. Usually, you can look at the few fractions and see it.

The most rigorous way is to write every equation:

(a)NaNO3 + (b)NaOH +(c) Zn -> (d)NH3 + (e)Na2ZnO2 + (f)H2O

Then do a balance for each element, ie:
Sodium:
a + b = 2e

Nitrogen:
a = d

Oxygen:
3a = 2e +f

etc ...

As a practical matter, zero in on the fact that all the hydrogen is in (b). And Zinc tells you (c) = (e). Since there are 5 hydrogens on the left, (b) HAS to be a multiple of 5.

As a "trick", just look for those elements that are most simply connected right and left, then try just try them. If you were to try the 5,10,15 series for (b) it would go quickly.

Like in simplifying a fraction after finding a common denominator, you want the final equation in the smallest denominator. If you add 1/2 + 1/3 + 1/4 you can either spot they all could have 12, or use 24 as the common denominator ... 12/24 + 8/24 + 6/24 =15/24 = 26/24 =13/12

But the main trick is to spot the elements that are simply connected, and then plug those in to get the other elements.

[Mentor Note: explicit solution removed]

Generally, I am looking for the easiest integer relationships, then trying integer multiples of those ratios until I spot one. I find that trial and error approach fastest.

 

[votingmachine]

Sorry to slip across the line. The "trick" is that trial-and-error is a valid and easy approach. It is how I always approach it. It has been too long since I learned to remember how it was taught to me.

It is somewhat like successive approximations. You want to start with an informed starting point by noticing something. But you can pretty quickly run thru some integers and see how they affect the balance of the elements left and right, then tweak it for another trial.

I think trial-and-error is sometimes overlooked by students who are trained to "solve" the problem. You are looking for a set of integers, and the relationship between them will often fall out simply by trying a few. Again, try to spot the best starting point ... but starting is more important.

I don't know if that is a "trick" or not.

 

[Borek]

The most rigorous way is to write every equation:

(a)NaNO3 + (b)NaOH +(c) Zn -> (d)NH3 + (e)Na2ZnO2 + (f)H2O

Then do a balance for each element, ie:
Sodium:
a + b = 2e

This is so called algebraic method. Yes, it is the most rigorous and general method, no, it is not failproof.

In general it produces infinitely many solutions, so requires one additional condition: all coefficients must be the smallest possible integers.

Even then it can occasionally fail, but these are extremally rare cases.