If you add 97 molecules of H2SO4 into a beaker and add 55 molecules of NaOH to same beaker, how many total molecules of water after the reaction?

CONCEPT INVOLVED
Sulfuric acid(H₂SO₄) is a strong acid and Sodium Hydroxide(NaOH) is a strong base.When they are mixed,they react with each other.It is an acid -base neutralization reaction between a strong acid and a strong base.
The products will be salt and water.The salt formed here is sodium sulphate(Na₂SO₄).
We can write the balanced chemical reaction as
H₂SO₄ + 2NaOH ---> Na₂SO₄ + 2H₂O + Heat
The reaction is highly exothermic as enormous amount of heat is liberated.
STEP 1
As we can see that 1 mole of H₂SO₄ is combining with 2 moles of NaOH to give 2 moles of water.
By using mole concept we can say that 1 mole of any entity = 6.022 x 10 23 entities (this big number is also called the Avogadro constant).
Thus we can say that 1 molecule of H₂SO₄ is combining with 2 molecules of NaOH to give 2 molecules of water.
STEP 2
In other words 1 molecule of H₂SO₄ need 2 molecules of NaOH to give 2 molecules of water.
Now we have 97 molecules of H₂SO₄ which will require 194(=97x2) molecules of NaOH.But we have only 55 molecules of NaOH.
Thus we can say that H₂SO₄ is in excess and it is NaOH which will be completely consumed and will act as a limiting reagent.
STEP 3
2 molecules of NaOH need 1 molecule of H₂SO₄ ( half the number of molecules) to give 2 molecules of water.
So 55 molecules of NaOH need 27.5 (=55 x 1/2) molecules of H₂SO₄.
2 molecules of NaOH need 1 molecule of H₂SO₄ to give 2 molecules of water.
Therefore,
55 molecules of NaOH need 27.5 molecules of H₂SO₄ to give 55 molecules of water.
(here number of molecules of water formed is equal to the number of molecules of NaOH reacted)
Remaining molecules of H₂SO₄ =(97 - 27.5) =69.5 molecules will remain unreacted as there is no more molecules of NaOH to react with.
Hence if you add 97 molecules of H₂SO₄ into a beaker and add 55 molecules of NaOH to the same beaker you will get 55 molecules of water after the reaction is complete.

The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) can be written as follows:
H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O
Using stoichiometry, we can say that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide to generate 1 mole of sodium sulfate (Na₂SO₄) and 2 moles of water.
Also, remember that 1 mole of any substance contains 6.023 x 10^23 molecules of that substance.
This means that we can also state that 1 molecule of sulfuric acid reacts with 2 molecules of sodium hydroxide to generate 1 molecule of sodium sulfate and 2 molecules of water.
Here, we are given with 97 molecules of sulfuric acid, but only 55 molecules of sodium hydroxide. 97 molecules of sulfuric acid will need 194 (= 2 x 97) molecules of sodium hydroxide for complete reaction. This means that the sodium hydroxide amount will control the reaction, since it is in a limited amount.
This means that 55 molecules of sodium hydroxide will react with 27.5 (= 55 x 1/2) molecules of sulfuric acid, leaving 69.5 (= 97-27.5) molecules of sulfuric acid as unused.
From the balanced chemical reaction, we can see that 2 molecules of sodium hydroxide, upon complete reaction with sulfuric acid, generate 2 molecules of water. In other words, 1 molecule of sodium hydroxide will result in 1 molecule of water, and hence, 55 molecules of sodium hydroxide will result in the formation of 55 molecules of water.
Hope this helps.