Respuesta :
Answer: The limiting reagent is barium nitrate, the theoretical yield of barium sulfate is 3.03 grams and the percent yield of the reaction is 82.18 %.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
- For potassium sulfate:
Molarity of potassium sulfate solution = 1.76 M
Volume of solution = 19.9 mL
Putting values in equation 1, we get:
[tex]1.76M=\frac{\text{Moles of potassium sulfate}\times 1000}{19.9}\\\\\text{Moles of potassium sulfate}=\frac{1.76\times 19.9}{1000}=0.035mol[/tex]
- For barium nitrate:
Molarity of barium nitrate solution = 0.896 M
Volume of solution = 14.4 mL
Putting values in equation 1, we get:
[tex]0.896M=\frac{\text{Moles of barium nitrate}\times 1000}{14.4}\\\\\text{Moles of barium nitrate}=\frac{0.896\times 14.4}{1000}=0.013mol[/tex]
The chemical equation for the reaction of potassium sulfate and barium nitrate follows:
[tex]K_2SO_4+Ba(NO_3)_2\rightarrow BaSO_4+2KNO_3[/tex]
By Stoichiometry of the reaction:
1 mole of barium nitrate reacts with 1 mole of potassium sulfate
So, 0.013 moles of barium nitrate will react with = [tex]\frac{1}{1}\times 0.013=0.013mol[/tex] of potassium sulfate
As, given amount of potassium sulfate is more than the required amount. So, it is considered as an excess reagent.
Thus, barium nitrate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 moles of barium nitrate produces 1 mole of barium sulfate.
So, 0.013 moles of barium nitrate will produce = [tex]\frac{1}{1}\times 0.013=0.013moles[/tex] of barium sulfate.
Now, calculating the mass of barium sulfate by using the equation:
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.013 moles
Putting values in above equation, we get:
[tex]0.013mol=\frac{\text{Mass of barium sulfate}}{233.4g/mol}\\\\\text{Mass of barium sulfate}=(0.013mol\times 233.4g/mol)=3.03g[/tex]
- To calculate the percentage yield of barium sulfate, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of barium sulfate = 2.49 g
Theoretical yield of barium sulfate = 3.03 g
Putting values in above equation, we get:
[tex]\%\text{ yield of barium sulfate}=\frac{2.49g}{3.03g}\times 100\\\\\% \text{yield of barium sulfate}=82.18\%[/tex]
Hence, the limiting reagent is barium nitrate, the theoretical yield of barium sulfate is 3.03 grams and the percent yield of the reaction is 82.18 %.
