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

See below in bold.

Step-by-step 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 %.

Answer 2

The answer to your question is: 83.9 %

Step-by-step 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 %