Respuesta :
The question is incomplete, here is the complete question:
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
Answer: The percent yield of water in the reaction is 46.85 %.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For NaOH:
Given mass of NaOH = 77.0 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol[/tex]
- For sulfuric acid:
Given mass of sulfuric acid = 72.6 g
Molar mass of sulfuric acid = 98 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol[/tex]
The chemical equation for the reaction of NaOH and sulfuric acid follows:
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
So, 0.741 moles of sulfuric acid will react with = [tex]\frac{2}{1}\times 0.741=1.482mol[/tex] 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 produces 2 moles of water
So, 0.741 moles of sulfuric acid will produce = [tex]\frac{2}{1}\times 0.741=1.482moles[/tex] 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:
[tex]1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g[/tex]
To calculate the percentage yield of water, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of water = 12.5 g
Theoretical yield of water = 26.68 g
Putting values in above equation, we get:
[tex]\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%[/tex]
Hence, the percent yield of water in the reaction is 46.85 %.
