Question

Determine the theoretical yield of ZnS that can be prepared from the reaction of 0.488 g Zn with 0.503 g S8. What is the percent yield, if 0.606 g of ZnS were experimentally recovered

Answer 1

Answer: The percentage yield of zinc sulfide is 82.90 %

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(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:

`\text{Moles of zinc}=(0.488g)/(65.4g/mol)=0.0075mol`

• For sulfur:

Given mass of sulfur = 0.503 g

Molar mass of sulfur = 256 g/mol

Putting values in equation 1, we get:

`\text{Moles of sulfur}=(0.503g)/(256g/mol)=0.00196mol`

The given chemical equation follows:

`8Zn+S_8\rightarrow 8ZnS`

By Stoichiometry of the reaction:

8 moles of zinc reacts with 1 mole of sulfur

So, 0.0075 moles of zinc will react with =
`(1)/(8)* 0.0075=0.00094mol` 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 =
`(8)/(8)* 0.0075=0.0075moles` 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:

`0.0075mol=\frac{\text{Mass of zinc sulfide}}{97.5g/mol}\\\\\text{Mass of zinc sulfide}=(0.0075mol* 97.5g/mol)=0.731g`

To calculate the percentage yield of zinc sulfide, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of zinc sulfide = 0.606 g

Theoretical yield of zinc sulfide = 0.731 g

Putting values in above equation, we get:

`\%\text{ yield of zinc sulfide}=(0.606g)/(0.731g)* 100\\\\\% \text{yield of zinc sulfide}=82.90\%`

Hence, the percentage yield of zinc sulfide is 82.90 %