<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:
![\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20sulfuric%20acid%7D%3D%5Cfrac%7B72.6g%7D%7B98g%2Fmol%7D%3D0.741mol)
Given mass of NaOH = 77 g (Assuming)
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
![\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaOH%7D%3D%5Cfrac%7B77g%7D%7B40g%2Fmol%7D%3D1.925mol)
The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:
![H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O](https://tex.z-dn.net/?f=H_2SO_4%2B2NaOH%5Crightarrow%20Na_2SO_4%2B2H_2O)
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:
![1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g](https://tex.z-dn.net/?f=1.482mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B18g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20water%7D%3D%281.482mol%5Ctimes%2018g%2Fmol%29%3D26.67g)
- To calculate the percentage yield of water, we use the equation:
![\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Experimental yield of water = 12.5 g (Assuming)
Theoretical yield of water = 26.67 g
Putting values in above equation, we get:
![\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%20of%20water%7D%3D%5Cfrac%7B12.5g%7D%7B26.67g%7D%5Ctimes%20100%5C%5C%5C%5C%5C%25%20%5Ctext%7Byield%20of%20water%7D%3D46.9%5C%25)
Hence, the percent yield of water is 46.9 %