When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 1.50 g of copper was heated with excess sulfur to yield 1.59 g copper(I) sulfide. What is the percent yield

The percent yield of the reaction is 83.2 %.

Explanation:

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

     .....(1)

For copper:

Given mass of copper = 1.50 g

Molar mass of copper= 63.5 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of copper and sulphur follows:

Thus, copper is considered as a limiting reagent because it limits the formation of product and sulphur is in excess.

By Stoichiometry of the reaction:

2 moles of copper produces 1 mole of copper (I) sulphide

So,0.024 moles of copper will produce = of copper (I) sulphide

To calculate the percentage yield of copper (I) sulphide we use the equation:

Experimental yield of copper (I) sulphide = 1.59 g

Theoretical yield of copper (I) sulphide= 1.91 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 83.2 %.

im pretty sure its b