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?

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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...

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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.

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