Question

Use the following information to answer the following question.

Cu + 2 AgNO3 --> Cu(NO3)2 + 2 Ag

12.7 g of copper metal reacts with silver nitrate to actually produce 38.1 g of silver in the lab.

Percent Yield = Actual Yield/Theoretical Yield x 100%

What is the percent yield of silver (Ag) in this reaction?

Answer 1

For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%

Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver

12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:

38.1-g/43.2-g x 100% = 88.2%

Step-by-step explanation: