Chemistry 135
Clark College
Exam 2 Combined Key
You must show your work for full credit!
1. (Yellow) A solution contains both Ni(NO3)2 and Mg(NO3)2. Sodium hydroxide is added to the
solutions, and a white precipitate forms. Aqueous ammonia is added to the precipitate, and a blue
solution results, but a white precipitate remains.
a. What is the composition of the precipitate? Mg(OH)2
b. What is causing the blue color in the solution? Ni(NH3)22+ (just Ni2+ is green!)
(Purple) A solution contains both Ni(NO3)2 and Pb(NO3)2. Sodium hydroxide is added to the solutions,
and a white precipitate forms. Aqueous ammonia is added to the precipitate, and a blue solution
results, but a white precipitate remains.
c. What is the composition of the precipitate? Pb(OH)2
d. What is causing the blue color in the solution? Ni(NH3)22+ (just Ni2+ is green!)
2. Complete the reactions for the following sets of reagents (the reactions do not need to be
balanced). Be sure to include phase labels and charges, where appropriate. If no reaction occurs,
write NR.
Yellow
a. Zn(OH)2 (s) + xs OH-  Zn(OH)4-2 (aq)
b. Ag(NH3)2 + I-  AgI (s)
c. PbSO4 (aq) + heat  My error! I meant to indicate that the PbSO4 was a solid, and that
it would *not* dissolve with heat
Purple
d. Mg(OH)2 (s) + xs OH-  NR
e. PbSO4 + I-  PBI2 (s)
f.
AgSO4 (aq) + heat  My error again! I meant to indicate that the Ag2SO4 was a solid,
and that it would dissolve with heat.
3. (Yellow) A solution containing an unknown anion is treated with lead (II) nitrate, and a bright yellow
precipitate forms. What was the unknown anion?
That bright yellow precipitate is PbI2, so the unknown anion is I- (iodide).
(Purple) A solution containing an unknown cation is treated with sodium iodide, and a bright yellow
precipitate forms. What was the unknown cation?
That bright yellow precipitate is PbI2, so the unknown cation is Pb2+ (lead (II)).
Exam 2
Spring 2006
Page 1 of 2
Chemistry 135
Clark College
4. (Yellow) Ima Student prepares a different cobalt complex with the formula Co(NH3)3Cl3. 9.80 g of
CoCl2•2 H2O is mixed with 9.62 g of NH4Cl. Enough extra NaCl to provide enough chloride ions.
Ima isolates 5.64 g of the complex. What is the percent yield of this reaction?
Molecular weights: Co(NH3)3Cl3 = 216.4 g/mol; CoCl2•2 H2O = 165.9 g/mol; NH4Cl = 53.49 g/mol.
You must show your work for both NH4Cl and CoCl2•2 H2O in the limiting reagent process! The
stoichiometric ratios are 1:1 for Cobalt and 3:1 for ammonia. No balanced equation is needed –
the ratios can be determined from the complex formula.
Yellow:
9.80 g CoCl 2 •2 H2O x
9.62 g NH4Cl x
1 mol Co(NH3 )3Cl 3
mol
216.4 g
x
x
= 12.8 g complex
165.9 g
1 mol CoCl 2 •2 H2O
mol
1 mol Co(NH3 )3Cl 3
mol
216.4 g
x
x
= 13.0 g complex
53.49 g
3 mol NH4Cl
mol
Cobalt is the limiting reagent, the theoretical yield is 12.8 g.
5.64 g
% yield =
x 100% = 44.1%
12.8 g
Purple:
11.75 g CoCl 2 •2 H2O x
11.59 g NH4Cl x
1 mol Co(NH3 )3Cl 3
mol
216.4 g
x
x
= 15.33 g complex
165.9 g
1 mol CoCl 2 •2 H2O
mol
1 mol Co(NH3 )3Cl 3
mol
216.4 g
x
x
= 15.64 g complex
53.49 g
3 mol NH4Cl
mol
Cobalt is the limiting reagent, the theoretical yield is 15.33 g.
8.92 g
% yield =
x 100% = 58.2%
15.33 g
5. Determine the number of ions that would form in solution when the following compounds are
dissolved in water.
Complex
Number of Ions
Complex
Number of Ions
Na[Mn(CN)3Br3]
2
K2[NiCl4]
3
[Fe(H2O)6]2(SO4)3
5
[Fe(H2O)6](ClO4)3
4
[Rh(NH3)2Cl4]Cl•H2O
2
Na[Mn(CN)3Br3]
2
K2[NiCl4]
3
[Rh(NH3)2Cl4]Cl•H2O
2
Waters of hydration (the •H2O) are not ions!
Exam 2
Spring 2006
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