Question

1. In a lab, you will run 3 reactions, 2 of which will result in the formation of the product listed below.

Reaction 2: 10 mL of 0.1 M CuSO4 and 10 mL of 0.1 M NaOH

Reaction 3: 10 mL of 0.1 M Pb(NO3)2 and 10 mL of 0.1 M CuSO4

a. Write the full ionic and the net ionic equation for the reaction. Identify the spectator ions.

b. Identify the limiting reagent for the reaction. If there is no limiting reagent, explain why.

c. Calculate the theoretical yield in grams of the product.

d. Calculate the amount of excess reagent left at the end of the reaction in grams.

e. In a box, draw a model of the species present in the beaker at the end of each reaction. Be sure to identify the parts of the model.

Answer 1

a)

1) balanced equation : 2 NaOH(aq) + CuSO4(aq) ---------> Na2SO4(aq) + Cu(OH)2(s)

Ionic equation: 2Na+ + 2OH- + Cu2+ + SO4^2- ---------------> 2Na+ + SO4^2- + Cu(OH)2(s)

Net ionic: Cu2+ + 2OH- --------------> Cu(OH)2(s)

2)

Molecular equation : Pb(NO3)2(aq) + CuSO4(aq) ------------> PbSO4(s) + Cu(NO3)2(aq)

Full ionic equation : Pb2+(aq) + 2NO3 - (aq) + Cu2+(aq) + SO4 2-(aq) -------> PbSO4(s) + Cu2+(aq) + 2NO3-(aq)

Net ionic equation : Pb 2+(aq) + SO4 2-(aq) -------------> PbSO4(s)

b)

moles of CuSO4 = 10 x 0.1/1000 = 1 x 10^-3

moles of NaOH = 10 x 0.1 / 1000 = 1 x 10^-3

2 NaOH(aq) + CuSO4(aq) ---------> Na2SO4(aq) + Cu(OH)2(s)

   2                      1                           

1 x 10^-3         1x10^-3

here limiting reagent is NaOH.

2nd reaction :

moles of Pb(NO3)2 = 1 x 10^-3

moles of CuSO4 = 1 x 10^-3

Pb(NO3)2(aq) + CuSO4(aq) ------------> PbSO4(s) + Cu(NO3)2(aq)

   1                         1

1 x 10^-3             1 x 10^-3

here there is no limiting reagent. because moles of two reactants are same. so both consumed.

c)

2 NaOH(aq) + CuSO4(aq) ---------> Na2SO4(aq) + Cu(OH)2(s)

   2                      1                           

1 x 10^-3         1x10^-3

here limiting reagent is NaOH.so product formed based on that.

2 mol NaOH -----------> 1 mol Cu(OH)2

1 x 10^-3 mol NaOH ----------->   ??

moles of Cu(OH)2 = 5 x 10^-4 moles.

moles of Na2SO4 = 5 x 10^-4 moles