Respuesta :
Answer: The theoretical yield of barium sulfate is 50.9 grams
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 sodium sulfate:
Given mass of sodium sulfate = 32.4 g
Molar mass of sodium sulfate = 142 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol[/tex]
- For barium chloride:
Given mass of barium chloride = 65.3 g
Molar mass of barium chloride = 208.23 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol[/tex]
The chemical equation for the reaction of barium chloride and sodium sulfate follows:
[tex]Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl[/tex]
By Stoichiometry of the reaction:
1 mole of sodium sulfate reacts with 1 mole of barium chloride
So, 0.228 moles of sodium sulfate will react with = [tex]\frac{1}{1}\times 0.228=0.228mol[/tex] of barium chloride
As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sodium sulfate produces 1 mole of barium sulfate.
So, 0.228 moles of sodium sulfate will produce = [tex]\frac{1}{1}\times 0.228=0.228moles[/tex] of barium sulfate
Now, calculating the mass of barium sulfate from equation 1, we get:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.228 moles
Putting values in equation 1, we get:
[tex]0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g[/tex]
Hence, the theoretical yield of barium sulfate is 50.9 grams
