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

The percentage yield of zinc sulfide is 82.90 %

Explanation:

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

    .....(1)

Given mass of zinc = 0.488 g

Molar mass of zinc = 65.4 g/mol

Putting values in equation 1, we get:

Given mass of sulfur = 0.503 g

Molar mass of sulfur = 256 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

8 moles of zinc reacts with 1 mole of sulfur

So, 0.0075 moles of zinc will react with = 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 = 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:

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

Experimental yield of zinc sulfide = 0.606 g

Theoretical yield of zinc sulfide = 0.731 g

Putting values in above equation, we get:

Hence, the percentage yield of zinc sulfide is 82.90 %

points

explanation:

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