how many grams of Ag will be produced from 19.0 g of Cu and 125 g of AgNO3

Question

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

Answer 1

Step 1.) Find limiting reactant after converting reactants to moles.

Mass Given / (Molar Mass) = Moles

19.0g Cu / 63.55 g Cu = 0.299 mol Cu

125g AgNO₃ / (107.87g + 14.00g + 3(16.00g)) AgNO₃ = 0.736 mol AgNO₃

I can find the limiting reactant by picking one of the reactants and converting it into the other to find out how much of the other reactant is consumed. I convert using stoichiometry and the mol ratio given by the balanced equation [ Cu + 2AgNO₃ = 2Ag + Cu(NO₃)₂ ]

0.299 mol Cu x (2 mol AgNO₃ / 1 mol Cu) = 0.598 mol AgNO₃ required to react 0.299 mol Cu

Since I have 0.736 mol AgNO₃ and only needed 0.598 mol AgNO₃, that means AgNO₃ is in excess. Therefore, Cu is my limiting reactant!

Step 2.) Use limiting reactant to calculate amount of moles of Ag that form

0.299 mol Cu x ( 2 mol Ag / 1mol Cu) = 0.598 mol Ag

Step 3.) Convert moles of Ag to g of Ag using Molar Mass of Ag(107.87g)

0.598 mol Ag x (107.87 g/ 1mol Ag) = 64.5 g Ag