Limiting Reactants: Comparison of Moles Method

Question

For the following reaction, 147 grams of silver nitrate are allowed to react with 23.1 grams of copper .

silver nitrate(aq) + copper(s)  copper(II) nitrate(aq) + silver(s)

What is the maximum amount of copper(II) nitrate that can be formed? 

What is the formula for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? 

Answer 1

For me, the easiest way to find the limiting reactant is to find the moles of each reactant, divided that number by the coefficient in the balanced equations and compare results. In the present problem, it would look like this:

2AgNO₃(aq) + Cu(s) ==> Cu(NO₃)₂(aq) + 2Ag(s) ... balanced equation

mols AgNO₃ = 147 g x 1 mol/169.9 g = 0.865 mols (÷2 -> 0.433)

mols Cu = 23.1 g x 1 mol/63.6 g = 0.363 mols (÷1 -> 0.363)

Since 0.363 is less than 0.433, Cu IS LIMITING

To find theoretical yield, use the mols of the limiting reactant and find mass of product:

0.363 mols Cu x 1 mol Cu(NO₃)₂ / 1 mol Cu x 187.6 g/mol = 68.1 g Cu(NO₃)₂

To find amount of excess reagent that remains, find out how much was used, and subtract that from what you started with:

mols AgNO₃ used = 0.363 mols Cu x 2 mols AgNO₃ / mol Cu = 0.726 mols AgNO₃ used up

mols AgNO₃ initially present = 0.865 mols (see calculation above when looking at limiting reactant)

mols AgNO₃ left over = 0.865 - 0.726 = 0.139 mols AgNO₃ remaining

mass AgNO₃ left over = 0.139 mols AgNO₃ x 169.9 g/mol = 23.6 g AgNO₃ remaining