4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

Question

If a container were to have 16 molecules of NH3 and 16 molecules of O2 initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion? (Use only whole numbers to solve.)

Answer 1

I'm assuming from the statement about whole numbers that we don't "count" parts of a molecule.
What's the limiting reagent(LR)?
16 NH3 x (4 NO/4 NH3) = 16 x 4/4 = 16 molecules NO formed if we had 16 molecules NH3 and all the O2 we needed.

16 O2 x (4 NO/5 O2) = 16 x 4/5 = 12.8 molecules NO formed if we had 16 molecules O2 and all the NH3 we needed.

Therefore, O2 is the LR and NH3 is the excess reagent (ER).

Next we calculate molecules NO and H2O formed from the O2.
For NO:
16 O2 x (4 NO/5 O2) = 12.8 molecules NO formed. Since we are to use ONLY whole numbers, this is 12 molecules NO and we used 15 of the 16 molecules O2 which leaves 1 molecule O2 unused. You can check that out by
15 O2 x 4/5 = 12 molecules NO formed. The 16 O2 - 15 O2 = 1 molecule O2 left unreacted.

For H2O:
We're using only 15 of the O2 now so
15 O2 x (6 H2O/5 O2) = 15 x 6/5 = 18 molecules. That's 18 molecules H2O formed (of course with the 1 molecule O2 left over but that's the same molecule O2 we had left over from the NO calculation).

How much NH3 is used:
15 O2 x (4 NH3/5 O2) = 15 x 4/5 = 12 molecules NH3 used. We had 16 to begin with so we have 4 molecules NH3 left over.
Total molecules in the container now is
12 NO formed
1 O2 unreacted
18 H2O formed
4 NH3 unreacted
total is 12 + 1 + 18 + 4 = ?