<u>Answer:</u> The percent yield of water is 46.9 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For sulfuric acid:</u>
Given mass of sulfuric acid = 72.6 g (Assuming)
Molar mass of sulfuric acid = 98 g/mol
Putting values in equation 1, we get:

Given mass of NaOH = 77 g (Assuming)
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

By Stoichiometry of the reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
0.741 moles of sulfuric acid will react with =
of NaOH
As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.
Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sulfuric acid reacts with 2 moles of water
0.741 moles of sulfuric acid will react with =
of water
- Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 1.482 moles
Putting values in equation 1, we get:

- To calculate the percentage yield of water, we use the equation:

Experimental yield of water = 12.5 g (Assuming)
Theoretical yield of water = 26.67 g
Putting values in above equation, we get:

Hence, the percent yield of water is 46.9 %