Answers and Explanations
Reactions and Periodicity 83- D—The balanced equation is:
- B—Carbonates produce carbon dioxide gas in
the presence of an acid. - D—Aluminum hydroxide, Al(OH) 3 , forms ini-
tially but then dissolves to form the Al(OH) 4 −ion. - 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. All the others are colorless. - B—Lead(II) carbonate is insoluble, so its
formula should be left 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. - C—The balanced chemical equation is:
The copper is below hydrogen on the activity
series, so H 2 cannot be formed by this
acid–metal reaction. Nitric acid is an oxidizing
agent, which will oxidize copper to Cu^2 +giving
Cu(NO 3 ) 2.- A—Acetic acid is a weak acid; as such it should
appear as HC 2 H 3 O 2. Potassium hydroxide is
a strong base so it will separate into K+ and
OH−ions. The potassium ion is a spectator ion,
and is left out of the net ionic equation. - E—Aqueous ammonia contains NH 3. The reac-
tion produces the silver–ammonia complex,
[Ag(NH 3 ) 2 ]+. - A—The reaction is:
KOH is a water-soluble strong base, not an acid.
As a strong base it will react with an acid.
Iron(II) hydroxide, Fe(OH) 2 , is insoluble and
will precipitate.- E—Aqueous ammonia contains NH 3. The
charges on each side of the reaction arrow must
be equal. - B—Neither B nor C is soluble in water. Only B
will react will nitric acid, and will dissolve. - 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 +. - D—The HCl will react with one-half the silver
to halve the concentration. The doubling of
the volume halves the concentration a second
time. - B—All the potassium and nitrate ions remain in
solution. However, two nitrate ions are pro-
duced per solute formula as opposed to only one
potassium ion. The lead and potassium would be
equal, but some of the lead is precipitated
as PbCl 2. - C—Ammonia, as a base, will precipitate the
metal hydroxides. Chromate, sulfide, and
chloride ions will precipitate one or more of the
ions. - E—Chlorine is an oxidizing agent. It is capable
of oxidizing both B and E. Answer B gives I 2 ,
which is brownish in water and purplish in
methylene chloride. Answer E gives reddish Br 2. - D—The following reaction occurs: NH 4 +(aq) +
OH−(aq) →NH 3 (g) +H 2 O(l) - B—The precipitate is PbCrO 4.
22 KHO KOHH+→ + 222
38 3
24
332
2Cu HNO Cu NO
NO H O+→ +
+
()
32
6
234
342 2Fe OH s H PO aq
Fe PO s H O l