Cu+2AgNO3→2Ag+Cu(NO3)2 BCA table

Question

If I have 33.4 g of copper reacting with 112.4 g of AgNO3, how many grams of Ag form?

Is Cu the limiting factor since it has less mols?

Answer 1

Hello, Dawn,

Cu+2AgNO₃→2Ag+Cu(NO₃)₂

Your guess that Cu would be the limiting reactant sounds reasonable, but look at what the equation is telling us. The masses alone do not say how many molecules we have of both. But we can use moles to determine that number.

The equation says we need 2 silver nitrate molecules for every one copper atom. So let's calculate how many moles we have of both reactants and see if there are enough silver nitrate molecules. I get:

Cu: 33.4grams/63.55 gramsmole = 0.526 moles

AgNO₃: 112.4 grams/170 grams/mole = 0.662 moles

I see a problem - if we consumed all of the copper, we would need 2x0.526 moles of silver nitrate. That's over a mole, and we have well under that. So it's the silver nitrate that is the limiting reagent. If we consume all of the AgNO₃ we only need 1/2 that number of moles of copper, or 0.33 moles. That will leave 0.331 moles of the copper unreacted.

To answer the question of how many grams we get of the silver, use the molar ratio of Ag to AgNO₃ of 2 to 2, or 1 to 1. We'll get the same number of moles of Ag as were consumed of the AgNO₃. Since we consumed all of the AgNO₃, 0.662 moles, we'll get the same number of moles of the silver. Multiply that by the molar mass of silver to find grams Ag: I get 71.4 grams of silver. Remove the excess copper and head for a jeweler.

Bob