Download First assignment 1996

Chemistry 1000 A
Fall 2004
Answers to Problem Set #3
1.
Balance the following equations, and name the reaction products:
(a) UO2(s) + 4 HF(l) → UF4(s) + 2 H2O(l): Uranium tetrafluoride and water (dihydrogen oxide, if you wish!)
(b) 5 NH3(aq) + 3 HNO3(aq) → 4 N2(g) + 9 H2O(l): Dinitrogen and water. NB: this is a redox reaction, and is rather
difficult to balance by inspection. The key to balancing is recognizing the change in oxidation states of the N atoms in
the two reactants.
(c) BF3(g) + 3 H2O(l) → 3 HF(aq) + H3BO3(aq): Hydrofluoric acid and boric acid. NB: the structure of boric acid is
actually B(OH)3, which makes it look like it should be a hydroxide. In fact, it is a very weak acid and some of the
hydrogen atoms ionize to H+.
2.
Which compound or compounds in each of the following groups is (are) expected to be soluble in water? If it is soluble,
state which ions are produced in water.
(a) PbSO4, Pb(NO3)2, and PbCO3: Pb2+(aq) and NO3–(aq) ions.
(b) Na2SO4, NaClO4, and NaCH3CO2: Na+ (aq), SO42–(aq) , ClO4–(aq) , and CH3CO2–(aq)
(c) AgBr, KBr, Al2Br6: K+(aq), Al3+aq), Br–(aq)
3.
Phosphoric acid can supply one, two, or three H+ ions in aqueous solution. Write balanced equations to show this
successive loss of hydrogen ions.
1)
H3PO4 + H2O → H3O+ + H2PO4–
2)
H2PO4– + H2O → H3O+ + HPO42–
3)
HPO42– + H2O → H3O+ + PO43–
4.
Balance each of the following equations, and then write the net ionic equation:
(a) (NH4)2CO3(aq) + Cu(NO3)2(aq) → CuCO3(s) + 2 NH4NO3(aq)
Net Ionic: CO32–(aq) + Cu2+(aq) → CuCO3(s)
(b) Pb(OH)2(s) + 2 HCl(aq) → PbCl2(s) + H2O(l)
Net Ionic: Pb(OH)2(s) + 2 H+ + 2 Cl– (aq) → PbCl2(s) + H2O(l)
(c) BaCO3(s) + 2 HCl(aq) → BaCl2(aq) + H2O(l) + CO2(g)
Net Ionic: BaCO3(s) + 2 H+ (aq) → Ba2+(aq) + H2O(l) + CO2(g)
(d) 4 HCl(aq) + MnO2(s) → MnCl2(aq) + Cl2(g) + 2 H2O(l)
Net Ionic: 4 H+(aq) + 2 Cl–(aq) + MnO2(s) → Mn2+(aq) + Cl2(g) + 2 H2O(l)
5.
Balance each of these equations and then classify each one as an acid-base reaction, a precipitation, or a gas-forming
reaction. Justify your classification:
(a) Ba(OH)2(s) + 2 HCl(aq) → BaCl2 + 2 H2O Precipitation reaction (BaCl2 is an insoluble combination of ions.) You
could also classify this reaction as an acid-base reaction between HCl and the hydroxide, but is is essential to
recognize that barium(II) chloride is insoluble.
(b) 2 HNO3(aq) + CoCO3(s) → Co(NO3)2 + H2O + CO2 Gas forming. CO2 is a gas. Again, it could be seen as an acid
base reaction, as carbonate ion is considered to be basic, but it is not a hydroxide base. Choose gas forming!
(c) 2 Na3PO4(aq) + 2 Cu(NO3)2(aq) → Cu3(PO4)2 + 6 NaNO3
Precipitation reaction. Cu3(PO4)2 is insoluble. This is
a classic example of a metathesis reaction, also known as a double-displacement reaction. It is driven forward
using Le Chaterlier’s principle by the insolubility of cupric phosphate.
(e) Fe(OH)3(s) + 3 HNO3(aq) → Fe(NO3)3 + 3 H2O
Acid-base. The metal hydroxide reacts with nitric acid. The
products are fully soluble, so this is the only classification that fits.
(f) FeCO3(s) + 2 HNO3(aq) → Fe(NO3)2 + CO2 + H2O
Gas forming. See comment to (b).
(g) FeCl2(aq) + (NH4)2S(aq) → FeS + 2 NH4Cl Precipitation, since iron(II) sulfide is an insoluble combination of ions.
(h) Fe(NO3)2(aq) + Na2CO3(aq) → FeCO3 + 2 NaNO3
ions.
Precipitation. Iron(II) carbonate is an insoluble combination of
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6.
Complete and balance the following acid-base reactions. Name the reactants and products.
Phosphoric acid. Potassium hydroxide. Potassium dihydrogen
(a) H3PO4(aq) + KOH(aq) → KH2PO4(aq) + H2O(l)
phosphate. Water.
Phosphoric acid. Potassium hydroxide. Potassium
or: H3PO4(aq) + 2 KOH(aq) → K2HPO4(aq) + 2 H2O(l)
hydrogen phosphate. Water.
Phosphoric acid. Potassium hydroxide. Potassium
or: H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)
phosphate. Water.
Any one of these three answers is acceptable as you are not told what ratio the reactants are to be mixed in!
(b) H2C2O4(aq) + Ca(OH)2(s) → CaC2O4(s) + 2 H2O(l) (H2C2O4 is oxalic acid, an acid capable of donating two H+
ions.)
Here you are told that oxalic acid can be doubly-deprotonated. However, it is actually a very tricky question. Both
oxalic acid and calcium hydroxide are weak. Normally, we would not expect them to react in this fashion. In fact,
the forward direction is driven by LeChatelier’s principle through the insolubility of calcium oxalate. This salt is
responsible for the dry taste associated with eating rhubarb or spinach. Both these vegetables contain high
concentrations of oxalic acid, and they react with calcium ions in the eater’s mouth to deposit a film of calcium
oxalate. High doses of oxalic acid can be toxic when calcium ions elsewhere in the body (e.g. in the blood stream)
are sequestered and rendered unavailable for essential processes such as nerve synapse responses.
7.
Many minerals are metal carbonates, and siderite is a mineral that consists largely of iron(II) carbonate. Write an
overall, balanced equation for the reaction of the main constituent of the mineral with sulfuric acid, and name each
reactant and product.
+
H2SO4(aq) → FeSO4(aq) + H2O(l) + CO2(g)
FeCO3(s)
iron(II) carbonate sulfuric acid
iron(II) sulfate water carbon dioxide
8.
Determine the oxidation number of each element in the following ions or compounds:
S +6
F –1
(a) SF6
H +1
As +5 O –2
(b) H2AsO4–
C0
H +1
O –2
(c) C2H4O2
N +4
O –2
(d) N2O4
(e) C5H8O2
C –0.8 H +1
O –2
NB: the carbon oxidation state is an average over several atoms.
Xe +6 O –2
(f) XeO42–
9.
Which of the following reactions are oxidation-reduction reactions? Explain your answer in each case. Classify the
remaining reactions.
Redox. Zn→Zn2+; N+5→N+4
(a) Zn(s) + 2 NO3–(aq) + 4 H+(aq) → Zn2+(aq) + 2 NO2(g) +2 H2O(l)
Acid-Base
(b) Zn(OH)2(s) + H2SO4(aq) → ZnSO4(aq) + 2 H2O(l)
Redox and gas-forming! Ca→Ca2+; H+1→0
(c) Ca(s) +2 H2O(l) → Ca(OH)2(s) + H2(g)
10. In each of the following reactions, tell which reactant is oxidized and which is reduced. Designate the oxidizing agent
and reducing agent.
(a) Ca(s) + 2 HCl(aq) → CaCl2(aq) + H2(g)
Ca is oxidized. It is the reducing agent. H+ is reduced. It is the oxidizing agent.
(b) Cr2O72–(aq) + 3 Sn2+(aq) + 14 H+(aq) → 2 Cr3+(aq) + 3 Sn4+(aq) +7 H2O(l)
Cr is reduced. It is part of the oxidizing agent, Cr2O72–(aq). Tin is oxidized. Sn2+(aq) is the reducing agent.
(c) FeS(s) + 3 NO3– (aq) + 4 H+(aq) → 3 NO(g)+SO42–(aq) + Fe3+(aq) +2 H2O(l)
Fe is oxidized. It is part of the reducing agent is FeS(s). N is reduced. It is part of the oxidizing agent NO3–(aq).
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11. Aspirin (C9H8O4) is produced by the reaction of salicylic acid (C7H6O3) and acetic anhydride (C4H6O3):
C7H6O3 (s) + C4H6O3 (l) � C9H8O4 (s) + CH3COOH (l)
If you mix 100. g of each of the reactants, what is the maximum mass of aspirin that can be obtained?
MM (g/mol)
C7H6O3 (s)
138.12
+
C4H6O3 (l)
102.089
�
C9H8O4 (s)
180.160
mass before rxn. (g) 100.
100.
0
mol before rxn.
0.7240
0.9795
0
mol after rxn.
0
(0.9795 – 0.7240)
0.7240
mass after rxn. (g)
+
CH3COOH (l)
130.43
NOTE: this is a limiting reagent problem, with reactants in a 1:1 mole ratio. We can therefore see immediately
without further calculation that the acetic anhydride is in excess, so that the salicylic acid is the limiting reagent in
this problem. All further calculations are therefore to be based on the 0.7240 moles of salicylic acid available. Thus
we can answer the problem:
The maximum yield of aspirin in this synthesis will be 130. g (to 3 s.f.)
Alternate: 1.30 · 102 g
12. An unknown metal reacts with oxygen to give the metal oxide, MO2. Identify the metal based on the following infor­
mation:
Mass of metal = 0.356 g
Mass of sample after converting metal completely to oxide = 0.452 g
+
M (s)
?
mass before rxn (g)
0.356
= 0.452 – 0.356 (Note this can be worked out either for one O2 or two O)
= 0.096
0.452
3.0 · 10–3
3.0 · 10–3
mass after rxn. (g)
mol before rxn.
O2 (g)
31.998
�
MM (g/mol)
MO2 (s)
Therefore the MM of the metal must be 0.356 g ⁄ 3.0 · 10–3 mol = 120 – 10 g⁄mol to 2 s.f. The arithmetic answer is
actually 118.7 g⁄mol, so the book probably meant you to choose Sn as the answer. However, to two significant figures,
you can only say that it is one of Sn, Sb, Te or I. Of these, of course, only Sn is a true metal.
13. To analyze an iron-containing compound, you convert all the iron to Fe2+ in aqueous solution and then titrate the
solution with aqueous KMnO4 according to the following balanced, net ionic equation:
MnO4– (aq) + 5 Fe2+ (aq) + 8 H+ (aq) � Mn2+ (aq) + 5 Fe3+ (aq) + 4 H2O(l)
If a 0.598-g sample of the iron-containing compound re-quires 22.25 mL of 0.0123 M KMnO4 for titration to the
equivalence point, what is the weight percent of iron in the compound?
MnO4– (aq) +
MM (g/mol)
vol. titrated (mL)
22.25
Molarity (mol/L)
0.0123
mol reacted
0.27675 millimol
mass before rxn. (g)
5 Fe2+ (aq) + 8 H+ (aq)
55.847
� Mn2+ (aq) + 5 Fe3+ (aq) + 4 H2O(l)
5 · 0.27675 = 1.38375 mmol
= 1.3875 mmol · 10–3 mol/mmol · 55.847 g/mol = 7.7278 · 10–2 g
%comp = [7.7278 · 10–2 g ⁄ 0.598 g] · 100% = 12.9% to 3 s.f.
14. Aluminum bromide, a valuable laboratory chemical, is made by the direct reaction of the elements:
2 Al (s) + 3 Br2 (l) Æ Al2Br6 (s)
What is the theoretical yield in grams of Al2Br6 if 25.00 mL of liquid bromine (density = 3.100 g mL–1) and 12.5 g
aluminum metal are used? Is any aluminum or bromine left over when the reaction has gone to completion? If so,
what mass of which reactant remains?
Note: these answers are done in more detail than the problem solutions in the text. You do not need to do them is such
length and detail, but look carefully at what I have done if you have difficulty seeing how the texts gets its answers.
Al2Br6 (s)
Statement ↓ ⁄ Equation→
2 Al (s) +
3 Br2 (l) Æ
26.98
159.8
533.4
MM (g mol–1)
Volume (mL)
25.0
(m = d × V = 3.10 g mL–1 × 25.0 mL)
Mass (g) before rxn
12.5
77.5
0
Moles before rxn
0.4633
0.4850
0
Moles if all Al used up
0
?
0.2317 (mol. Al2Br6 = ½ mol. Al from equation)
?
0
0.1617 (mol. Al2Br6 = 1/3 mol. Br2 from equation)
Moles if all Br2 used up
The if statement leading to the least amount of product identifies the limiting reagent! So Br2 is limiting.
Moles of reagents consumed 0.3233
0.4850
(mol. Al req'd. = 2/3 mol. Br2 used)
Moles after rxn
0.1400
0
0.1617 (mol. Al left = 0.4633–0.3233)
Mass after rxn. (g)
3.78
0
86.2
15. Boron forms an extensive series of compounds with hydrogen, all with the general formula BxHy.
+ n O2 (g) Æ B2O3 (s) + H2O (g)
BxHy (s)
If 0.1615 g of BxHy gives 0.4217 g of B2O3 when burned in excess O2 , what is the empirical formula of BxHy? In a
separate experiment, the approximate molar mass of the same boron hydride was determined by osmometry to be 52.9
g mol–1. What is the molecular formula of BxHy?
Although this is a stoichiometry problem, it is best done in terms of moles and grams of elements since we want the
empirical and molecular formulae.
Moles of B = 2 × mol. B2O3 = 2 × 0.4217 g ⁄ 69.62 g mol–1 = 0.01211 mol
Mass of B = 0.01211 mol × 10.81 g mol–1 = 0.1310 g
∴ mass H = 0.1615 – 0.1310 = 0.0305
Moles of H = 0.0305 ⁄ 1.008 g mol–1 = 0.0303; ∴ ratio of H:B = 0.0303:0.01211 = 2.5:1 = 5:2
The empirical formula of the compound is ∴ B2H5, and the MM of B2H5 = 2 × 10.81 + 5 × 1.008 = 26.66 g mol–1.
We get the molecular formula from the ratio: 52.9 g mol–1 ⁄ 26.66 g mol–1 ≈ 2, ∴ B4H10.
16. A mixture of sodium carbonate and sodium hydrogen carbonate is treated with aqueous hydrochloric acid. The
unbalanced equations for the resulting reactions are:
Na2CO3 (s) + HCl (aq) Æ NaCl (aq) + CO2 (g) + H2O (l)
NaHCO3 (s) + HCl (aq) Æ NaCl (aq) + CO2 (g) + H2O (l)
You treat 9.450 g of a Na2CO3/ NaHCO3 mixture with an excess of aqueous HCl and isolate after evaporation 9.355 g
of NaCl. What is the weight percent of each substance in the mixture? What happens to the unreacted HCl upon
evaporation, and what does this suggest about the appropriate place to conduct this experiment?
This problem is similar to the ones done in lecture and in the text, but differs in that there are two rather than one
reactive component in the mixture. Consider first that the reactions need to be balanced:
Na2CO3 (s) + 2 HCl (aq) Æ 2 NaCl (aq) + CO2 (g) + H2O (l)
NaHCO3 (s) + HCl (aq) Æ NaCl (aq) + CO2 (g) + H2O (l)
It is the difference in stoichiometry of the two equations that is the key to this problem. But because there are two
unknowns, we need to use algebra, and define two equations. There are several ways to do this. I will choose to use
the mass of Na2CO3 and NaHCO3. Thus, let x = the mass Na2CO3 and of y = NaHCO3 in grams.
First we can say:
[1] x + y = 9.450 g
and thus y = 9.450 – x
3
Now we calculate the number of moles of NaCl:
Then we can say:
9.355g
= 0.1600 mol NaCl
58.451g ⋅ mol −1
[2] 0.1600 mol NaCl = 2 × mol. Na2CO3 + 1 × NaHCO3 from the stoichiometry of the reaction.
Now we need to homogenize the equations, and we can do so by replacing mol by mass ⁄ molar mass.
0.1600 = 2 ×
xg
yg
−1 + 1 ×
106.01g ⋅ mol
84.015g ⋅ mol −1
Now substitute from [1]
0.1600 = 0.018866 x + 0.1125 – 0.01190 x
0.1600 – 0.11250 = (0.018866 – 0.01190) x
Finally, the %composition is for Na2CO3
and
x = 6.822 g of Na2CO3
6.822 g
× 100 = 72.19 % and so NaHCO3 = 27.81%.
9.450 g
There is excess HCl in this reaction. As the solution evaporates, HCl(g) is released into the atmosphere.
Such reactions should always be conducted in a properly ventilated fume hood to preven the acidic vapours
from entering the work space, where they will condense in any moist areas to reform hydrochloric acid.
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