Respuesta :
Answer: The percentage yield of zinc sulfide is 82.90 %
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 Zinc:
Given mass of zinc = 0.488 g
Molar mass of zinc = 65.4 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of zinc}=\frac{0.488g}{65.4g/mol}=0.0075mol[/tex]
- For sulfur:
Given mass of sulfur = 0.503 g
Molar mass of sulfur = 256 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sulfur}=\frac{0.503g}{256g/mol}=0.00196mol[/tex]
The given chemical equation follows:
[tex]8Zn+S_8\rightarrow 8ZnS[/tex]
By Stoichiometry of the reaction:
8 moles of zinc reacts with 1 mole of sulfur
So, 0.0075 moles of zinc will react with = [tex]\frac{1}{8}\times 0.0075=0.00094mol[/tex] of sulfur
As, given amount of sulfur is more than the required amount. So, it is considered as an excess reagent.
Thus, zinc is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
8 moles of zinc produces 8 moles of zinc sulfide
So, 0.0075 moles of zinc will produce = [tex]\frac{8}{8}\times 0.0075=0.0075moles[/tex] of zinc sulfide
Now, calculating the mass of zinc sulfide from equation 1, we get:
Molar mass of zinc sulfide = 97.5 g/mol
Moles of zinc sulfide = 0.0075 moles
Putting values in equation 1, we get:
[tex]0.0075mol=\frac{\text{Mass of zinc sulfide}}{97.5g/mol}\\\\\text{Mass of zinc sulfide}=(0.0075mol\times 97.5g/mol)=0.731g[/tex]
To calculate the percentage yield of zinc sulfide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of zinc sulfide = 0.606 g
Theoretical yield of zinc sulfide = 0.731 g
Putting values in above equation, we get:
[tex]\%\text{ yield of zinc sulfide}=\frac{0.606g}{0.731g}\times 100\\\\\% \text{yield of zinc sulfide}=82.90\%[/tex]
Hence, the percentage yield of zinc sulfide is 82.90 %
