Question

What is the theoretical yield of CuS when 31.8g of Cu(s) is heated with 50.0g of S? (Asume only CuS is produced in the reaction) what is the percent yield of CuS if only 40.0g of CuS can be isolated from the mixture?

Answer 1

Answer:

The answer to your question is: 83.9 %

Explanation:

Data

Cu = 31.8 g

S = 50 g

CuS = 40 g

yield = ?

Equation

                     Cu +   S   ⇒  CuS

MW Cu = 64 g

MW S = 32 g

MW CuS = 96 g

Ratio (theoretical/experimental)

Experimental    50/31.8 = 1.57

Theoretical        32/64 = 0.5     limiting reactant Cu

                              64 g of Cu ------------------  96 g of CuS

                              31.8 g        -------------------    x

                              x = (31.8 x 96) / 64

                             x = 47.7 g of CuS

% yield = (40/47.7) x 100

            = 83.9 %

                         

Answer 2

Answer:

See below in bold.

Explanation:

The balanced equation is:

Cu + S --->  CuS

Taking  relative atomic masses

63.546 g Cu reacts with 32.06 g of Sulfur to give 95.606 g CuS.

So the theoretical yield for the given masses is

(95.606 / 63.546) * 31.8 = 48.235 g of CuS.

Percent yield = (40 / 48.235) * 100

= 82.93 %.