Question

In a particular reaction between copper metal and silver nitrate, 12.7 g cu produced 38.1 g ag. what is the percent yield of silver in this reaction? cu + 2agno3 → cu(no3)2 + 2ag

Answer 1

88.4% !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

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%