For the following reaction, 108 grams of silver nitrate are allowed to react with 23.6 grams of copper . silver nitrate(aq) + copper(s) copper(II) nitrate(aq) + silver(s)

Question

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

Answer 1

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

Find limiting reactant:

mols AgNO3 = 108 g x 1 mol / 170 g = 0.635 mols AgNO3

mols Cu = 23.6 g x 1 mol / 63.6 g = 0.371 mols Cu

AgNO3 is the limiting reactant based on a mol ratio of 2 : 1 for AgNO3 : Cu.

Use the mols of the limiting reactant to find the theoretical yield of Cu(NO3)2:

0.635 mols AgNO3 x 1 mol Cu(NO3)2 / 2 mols AgNO3 x 188 g / mol = 59.7 g Cu(NO3)2

Excess reagent is Cu.

How much Cu is left after the reaction?

0.635 mols AgNO3 x 1 mol Cu / 2 mols AgNO3 = 0.3175 mols Cu used up

started with 0.371 mols Cu

mass Cu left over = 0.371 mols -0.3175 mols = 0.0535 mols left x 63.6 g/mol = 3.40 g Cu remaining