Answer: 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 %.