A reaction between copper metal and silver nitrate should theoretically produce 41.8 g Ag, however in the lab only 32.3 g Ag is actually produced. What is the percent yield of silver 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%

answer is a

explanation: