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?

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%

Explanation:

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