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 %

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.