Question

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?

Answer 1

Answer:

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 2

Answer is 88.2% :)) pls mark brainliest
Explanation: I got it right so trust me