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Chemistry 1010 Loader (Fall 2004)
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Chemistry 1010
Handout 4: Reactions, equations and Redox
1.
Write molecular and net-ionic (where applicable) equations
for the reactions that take place in the following experiments.
(a) Sodium hydroxide solution reacts with zinc nitrate
solution to give a precipitate of zinc hydroxide.
NaOH(aq) + Zn(NO3)2(aq) → Zn(OH)2(s) + NaNO3 (aq)
2 OH–(aq) + Zn2+(aq) → Zn(OH)2(s)
(b) Copper(II) oxide dissolves in dilute hydrochloric acid.
CuO(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l)
CuO(s) + 2 H+(aq) → Cu2+(aq) + H2O(l)
(c) Potassium chromate reacts with lead(II) nitrate solution
to give a yellow precipitate.
K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s) + 2 KNO3 (aq)
CrO42–(aq) + Pb2+(aq) → PbCrO4(s)
(d) Magnesium ribbon reacts with dilute hydrochloric acid.
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)
(e) Solid calcium carbonate reacts with dilute nitric acid to
give carbon dioxide gas.
CaCO3(s) + 2 HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + CO2(g) + H2O(l)
(f) Potassium metal reacts with cold water to give hydrogen
gas.
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)
2 K(s) + 2 H2O(l) → 2 K+(aq)+ 2 OH–(aq) + H2(g)
(g) Sodium hydrogen carbonate is heated strongly in a
crucible leaving a residue of sodium carbonate.
2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
(h) Potassium chlorate is heated strongly with a catalyst to
produce oxygen gas.
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
(i) P4O10(s) react and dissolves in water to give a solution
that turns blue litmus paper red.
P4O10(s) + 6 H2O(l) → 4 H3PO4(aq)
(j) Strontium oxide dissolves in water to give a solution that
turns red litmus paper blue.
(k) Aqueous barium nitrate reacts with aqueous lithium
sulfate to give a precipitate of solid barium sulfate and
aqueous lithium nitrate.
Ba(NO3)2(aq) + Li2SO4(aq) → BaSO4(s) + 2 LiNO3(aq)
Ba2+(aq) + SO42–(aq) → BaSO4(s)
(l) lead(II) nitrate solution is mixed with sodium sulfate
solution
Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2 NaNO3(aq)
SO42–(aq) + Pb2+(aq) → PbSO4(s)
(m) sodium hydroxide solution is added to a solution of
ammonium chloride (ammonia gas is evolved)
NaOH(aq) + NH4Cl(aq) → NaCl(aq) + NH3(g) + H2O(l)
OH–(aq) + NH4+(aq) → NH3(g) + H2O(l)
(n) calcium reacts with water
Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g)
Ca(s) + 2 H2O(l) → Ca2+(aq) + 2 OH–(aq) + H2(g)
(o) copper(II) sulfide(s) reacts with dilute sulfuric acid
CuS(s) + H2SO4(aq) → CuSO4(s) + H2S(g)
2 H+(aq) + S2–(aq) → H2S(g)
(p) aluminum sulfate solution is mixed with barium hydroxide
solution
Al2(SO4)3(aq) + 3 Ba(OH)2(aq) → 3 BaSO4(s) + 2 Al(OH)3(s)
2 Al3+(aq) + 3 SO42–+ 3 Ba2+ + 6 OH–(aq) → 3 BaSO4(s) + 2 Al(OH)3(s)
(q) lithium oxide(s) is added to water
Li2O(s) + H2O(l) → 2 LiOH(aq)
Li2O(s) + H2O(l) → 2 Li +(aq) + 2 OH–(aq)
(r) sodium bromide solution is added to an acidified (H2SO4)
potassium chromate solution
8 H2SO4(aq) + 6 NaBr(aq) + 2 K2CrO4(aq) → 3 Br2(aq) + Cr2(SO4)3(aq)
+ 8 H2O(l) + 2 K2SO4(aq) + 3 Na2SO4(aq)
16 H+(aq) + 6 Br–(aq) + 2 CrO42–(aq) → 3 Br2(aq) + 2 Cr3+(aq) + 8 H2O(l)
2.
What is the oxidation number of the UNDERLINED atom?
(a) HCl (b) HOCl
(c) S2O32– (d) As2O3
Na3AlCl6 (f) MnO42– (g) NH3 (h) Mg2P2O7
3.
4.
(e)
(a) -1 (b) +1 (c) +2 (d) +3 (e) +3 (f)+6
(g) -3 (h) +5
Write the formula for each of the following compounds:
(a) potassium sulfide; K2S
(b) barium chloride dihydrate; BaCl2.2H2O
(c) antimony(III) sulfide; Sb2S3
(d) calcium hydride; CaH2
(e) potassium chlorate; KClO3
(f) sulfurous acid; H2SO3(aq)
(g) calcium sulfite dihydrate; CaSO3.2H2O
(h) chromium(III) oxide; Cr2O3
Which of the following reactions is an oxidation-reduction
(REDOX) reaction. Write the NET-IONIC equation for each
reaction. Indicate where applicable, (i) the species being
oxidized and the species being reduced; and (ii) the oxidant
(oxidizing agent) and the reductant (reducing agent).
If REDOX reaction : Cu being reduced/oxidant S being
oxidized/reductant
(a) 2 Cu2O + Cu2S → 6 Cu + SO2
2 Cu2O + Cu2S → 6 Cu + SO2
(b) AgNO3(aq) + NaCl(aq) → AgCl(aq) + NaNO3(aq)
Not redox. Double replacement reaction
(c) 2 Al(s) + 3 F2(g) → 2 AlF3(s)
2 Al(s) + 3 F2(g) → 2 AlF3(s)
(d) 2 K2CrO4(aq) + H2SO4(aq) → K2Cr2O7(aq) + K2SO4(aq)
Not redox
(e) Cl2(aq) + 2 KI(aq) → I2(aq) + 2 KCl(aq)
Cl2(aq) + 2 KI(aq) → I2(aq) + 2 KCl(aq)
Na2O(s) + H2O(l) → 2 NaOH(aq)
Not redox
(g) 2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g)
2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g)
Calculate the oxidation number of the indicated element in
each of the following species.
(f)
5.
(a)
(b)
(c)
(d)
sulfur in thiosulfate ion; +2
nitrogen in ammonium chloride; -3
zinc in Na2ZnO2 ; +2
uranium in UO2(NO3)2; +2
Chemistry 1010 Loader (Fall 2004)
6.
(e) silicon in silicic acid (like carbonic acid); +4
(f) phosphorus in the fertilizer called triple superphosphate,
Ca(H2PO4)2; +5
Write balanced half-reactions for each of the following partial
half reactions.
(a) ClO2–(aq) → Cl2(g) in ACID solution
6 e– + 8 H+(aq) + 2 ClO2–(aq) → Cl2(g) + 4 H2O(l)
(b) S2O32– → S4O62– in ACID solution
2 S2O32– → S4O62– + 2 e–
(c) Cr(OH)3 → CrO42– in BASIC solution
5 OH – + Cr(OH)3 → CrO42– + 3 e– + 4 H2O(l)
7.
(d) N2H4 → N2 in BASIC solution
4 OH – + N2H4 → N2 + 4 e – + 4 H2O(l)
Balance the following redox equations in acid solution.
(a) MnO2(s) + PbO2(aq) → MnO4–(aq) + Pb2+(aq)
reduction PbO2(aq) → Pb2+(aq)
oxidation MnO2(s) → MnO4–(aq)
–
2 e + PbO2(aq) → Pb2+(aq) change in oxidation number = -2
MnO2(s) → MnO4–(aq) + 3 e– change in oxidation number = +3
2 MnO2(s) + 3 Pb–O2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq) :balance change
in ON
4 H+(aq) + 2 MnO2(s) + 3 PbO2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq)
:balance charges with H+
4 H+(aq) + 2 MnO2(s) + 3 PbO2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq) + 2
H2O(l) :balance H with H2O
BALANCED: 4 H+(aq) + 2 MnO2(s) + 3 PbO2(aq) → 2 MnO4–(aq) + 3
Pb2+(aq) + 2 H2O(l)
(b) Br2(aq) + SO2(g) → H2SO4(aq) + HBr(aq)
2 H2O(l) + Br2(aq) + SO2(g) → H2SO4(aq) + 2 HBr(aq)
8.
(d) Cr2O72–(aq) + I–(aq) → Cr3+(aq) + I2(s) (acid solution)
14 H+(aq) + Cr2O72–(aq) + 6 I–(aq) → 2 Cr3+(aq) + I2(s) + 7 H2O(l)
Concentrated sulfuric acid has a density of 1.84 g.cm–3 (1.00
mL = 1.00 cm3) and contains 96% of H2SO4 by weight.
(a) Calculate the concentration of the concentrated sulfuric
acid in mol.L–1. Ans: 18.0 mol L–1
(b) What volume of the concentrated sulfuric acid would be
required to make 2.50 litres of a 3.0 mol.L–1 solution of
the bench acid? Ans: 0.417 L
(c) What is the mass of the volume of the acid calculated in
(b)? Ans: 0.767 kg
9. (a) Calculate the concentration in mol.L–1 of a sodium
hydroxide solution 33.45 mL of which neutralizes 25.00
mL of 0.1500 mol.L–1 of nitric acid.
NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l)
mol HNO3 neutralized = 25.00 x 10–3 L x 0.1500 mol.L–1 =
3.750 x 10–3 mol. Since one mole of nitric acid reacts with
one mole of sodium hydroxide.. mole NaOH = 3.750 x 10–3
mol concentration of NaOH = 0.1121 mol.L–1
(b) A 10.00 mL aliquot of dilute sulfuric acid required 16.12
mL of 0.1021 mol.L–1 sodium hydroxide solution for
neutralization. What was the concentration of the sulfuric
acid solution?
2 NaOH(aq) + H2SO4(aq) → Na2SO4 (aq) + 2 H2O(l)
NaOH used = 16.12 x 10–3 L x 0.1021 mol.L–1 = 1.6459 x 10–3 mol
Since one mole of sulfuric acid reacts with two moles of
sodium hydroxide.
mole NaOH =
1
2
x 1.6459 x 10–3 mol = 8.2293 x 10–4 mol
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concentration of sulfuric acid = 0.08229 mol.L–1
(c) A 0.64753 g sample of the hydroxide of an unknown
metal (M) required 34.30 mL of 0.4244 hydrochloric acid
for complete reaction. The hydroxide is insoluble in water
and may have the formula MOH, M(OH)2 or M(OH)3. Use
the information from the titration to deduce the most
likely formula and hence the molar mass of the metal.
The possible reactions are
MOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
M(OH)2(aq) + 2 HCl(aq) → 2 NaCl(aq) + 2 H2O(l)
M(OH)3(aq) + 3 HCl(aq) → 3 NaCl(aq) + 3 H2O(l)
mol HCl used in the reaction = 34.30 x 10–3 L x 0.4244 mol.L–1
= 0.014557 mol
For the first equation one mole of acid reacts with one mole of
base so if MOH is the formula then the molar mass would be
0.64753 g
-1
0.014557 mol = 44.482 g mol giving about 44.48 - 17.00 =
37.48 g mol-1 for the molar mass of M
for M(OH)2(aq) the molar mass would be 2 x 44.482 g mol-1 =
88.965 g mol-1 giving about 89.87 - 34.00 = 55.87 g mol-1 for
the molar mass of M.
for M(OH)3(aq) the molar mass would be 3 x 44.482 g mol-1 =
133.45 g mol-1 giving about 133.45 - 51.00 = 82.45 g mol-1 for
the molar mass of M.
The is no metal with a molar mass close to 37.48 or 82.45 g
mol-1 with an insoluble hydroxide. However, for 55.87 g mol-1
this is almost exactly the molar mass of iron (55.85 g mol-1) so
the metal hydroxide is probably Fe(OH)2.
10. A 50.00 mL sample of a solution containing iron(II) ions,
Fe2+(aq), required 20.63 mL of a 0.0216 mol.L–1 solution of
KMnO4 for complete reaction according to the equation:
(incorrect wording in the original)
8 H+(aq) + MnO4–(aq) + 5 Fe2+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
(a) Calculate the concentration of iron(II) ions in the solution.
0.04456 mol L-1
(b) What volume of a 0.0150 mol.L–1 solution of KMnO4
solution would be required to react completely with the
iron(II) sulfate solution formed when 0.23245 g of iron
wire reacted with sulfuric acid?
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
mol iron =
0.23245 g
= 4.1620 x 10-3 mol
55.85 g mol -1
1
mol KMnO4 required = 5 x 4.1620 x 10-3 mol
1
-3
5 x 4.1620 x10 mol
so volume required =
= 5.549 x 10-–3 L
0.150 mol L -1
Ans: 5.549 mL
11. Balance the following equations in acid or base as indicated.
Identify the oxidant and the reductant.
(a) ClO3– + MnO4– → MnO2 + ClO4– (acidic)
2 H+ + 3 ClO3– + 2 MnO4– → 2 MnO2 + 3 ClO4– + H2O
(b) NO3– + S → NO + H2SO3 (acidic)
4 H+ + 4 NO3– + 3 S + H2O → 4 NO + 3 H2SO3
(c) Cr2O72– + I2 → IO3– + Cr3+ (acidic)
34 H+ + 5 Cr2O72– + 3 I2 → 6 IO3– + 10 Cr3+ + 17 H2O
(d) Zn + NO3– → ZnO22– + NH3 (basic)
7 OH–+ 4 Zn + NO3– → 4 ZnO22– + NH3 + 2 H2O
(e) S2– + ClO3– → Cl– + S (basic)
3 H2O + 3 S2– + ClO3– → Cl– + 3 S + 6 OH–