Reactions and Periodicity ‹ 85
❯ Answers and Explanations
- D—The balanced equation is 3 Fe(OH) 2 (s) +
2 H 3 PO 4 (aq) → Fe 3 (PO 4 ) 2 (s) + 6 H 2 O (l). - B—Carbonates produce carbon dioxide gas in
the presence of an acid. None of the other ions
will react with hydrochloric acid to produce a
gas. - A—Aqueous solutions of Cu^2 + are normally
blue. Iron ions give a variety of colors but are
normally colorless, or nearly so, in the absence of
complexing agents. The other ions are colorless. - B—Lead(II) carbonate is insoluble, so its for-
mula should be written as PbCO 3. Hydrochloric
acid is a strong acid, so it should be written as
separate H+ and Cl- ions. Lead(II) chloride,
PbCl 2 , is insoluble, and carbonic acid, H 2 CO 3 ,
quickly decomposes to CO 2 and H 2 O. Also
notice that A cannot be correct because the
charges do not balance. - C—The balanced chemical equation is 3 Cu(s) +
8 HNO 3 (aq) → 3 Cu(NO 3 ) 2 (aq) + 2 NO(g) +
4 H 2 O(l). The copper is below hydrogen on
the activity series, so H 2 cannot form by this
acid–metal reaction. Nitric acid causes oxida-
tion, which will oxidize copper to Cu^2 +, giving
Cu(NO 3 ) 2. Some of the nitric acid reduces
to NO. An oxidation and a reduction must
ALWAYS be together. - A—Acetic acid is a weak acid; as such, it should
be written as HC 2 H 3 O 2. Potassium hydroxide
is a strong base, so it will separate into K+ and
OH- ions. Any potassium compound that might
form is soluble and will yield K+ ions. The potas-
sium ions are spectator ions and are left out of
the net ionic equation. - A—Aqueous ammonia contains primarily NH 3 ,
which eliminates choices B and C. NH 3 Cl
does not exist, which eliminates choice D. The
reaction produces the silver-ammonia complex,
[Ag(NH 3 ) 2 ]^ +. - A—The reaction of potassium to produce a
potassium compound is an oxidation; therefore,
it must be a reduction, and the only species
available for reduction is hydrogen. The reaction
is 2 K(s) + 2 H 2 O(l) → 2 KOH(aq) + H 2 (g).
KOH is a water-soluble strong base, which will
not react with other strong bases.
9. B—The black color is due to the formation of
metallic mercury. Aqueous ammonia contains
primarily NH 3 , which eliminates choices C and
D. The total charges on each side of the reaction
arrow must be equal, which eliminates choice A.
10. D—The magnesium chloride gives 0.20 moles
of chloride ion, and the potassium chloride gives
0.10 moles of chloride ion. A total of 0.30 moles
of chloride will react with 0.15 moles of lead,
because two Cl- require one Pb^2 +.
11. D—The HCl is the limiting reagent. The HCl
will react with one-half the silver to halve the
concentration. The doubling of the volume
(50 mL + 50 mL) halves the concentration a
second time.
12. B—All the potassium and nitrate ions remain in
solution, leaving PbCl 2 as the only possible pre-
cipitate. Equal volumes of equal concentrations
give the same number of moles of reactants;
however, two nitrate ions are produced per
solute formula as opposed to only one potassium
ion. Initially, the lead and potassium would be
equal, but some of the lead is precipitated as
PbCl 2.
13. C—Ammonia, as a base, will precipitate the
metal hydroxides since the only soluble hydrox-
ides are the strong bases. Chromate, sulfide, and
chloride ions might precipitate one or more of
the ions. Nitrates, from nitric acid, are soluble;
therefore, this is the solution that is least likely
to cause an observable change.
14. D—Chlorine causes oxidation. It is capable
of oxidizing both B and D. Answer B gives I 2 ,
which is brownish in water and purplish in
methylene chloride. Bromine solutions are red-
dish in both.
15. D—Excess strong base will ensure the solution
is basic and not neutral. Both ammonium and
sodium salts are soluble; therefore, no precipi-
tate will form. One aqueous solution will mix
with another aqueous solution. The following
acid–base reaction occurs to release ammonia
gas: NH 4 +(aq) + OH-(aq) → NH 3 (g) + H 2 O(l).
16. B—The balanced equation is: 4 C 4 H 11 N(l) +
27 O 2 (g) → 16 CO 2 (g) + 22 H 2 O(l) + 2 N 2 (g).