Stoichiometry Problem

Question

In a particular reaction between copper metal and silver nitrate, 12.7 g Cu produces 38.1 g Ag. What is the percent yield of silver in this reaction?

                               Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

Answer 1

Hello, Lilia,

Thanks for including a correctly balanced reaction

Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

We can see that we would expect 2 moles of silver for every 1 mole of copper consumed. That's a molar ratio of 2 to 1. We will assume there is sufficient AgNO₃ to react with all of the copper. So if we calculate the moles of Cu in 12.7 grams of Cu, we can double it to find the moles of silver we'd expect.

Moles Cu: 12.7 g Cu/(63.5g/mole Cu) = 0.1999, or 0.200 moles Cu. Double that to find moles of Ag, 0.40 moles Ag. Multiply by silver's molar mass of 107.9 g/mole to find 43.11 grams of silver. That's the amount we'd expect if everything went perfectly.

We actually got 38.1 grams, so the yield is:

(38.1g/43.11g)*100% = 88.4% yield

Not bad.

Bob