Chemistry 3 & 4, Chapter 1 SOLUTIONS
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. In the following reaction, 6 mol of aluminium is mixed with 6 mol of oxygen:
4Al(s) + 3O2(g)
2Al2O3(s)
What amount of aluminium oxide will be produced from this reaction?
A 1 mol
B 2 mol
C 3 mol
D 4 mol
C
ANSWER:
FEEDBACK: From the equation, 6 mol of Al only requires 4.5 mol of O2, so O2 is in excess and Al
is the limiting reagent. Since
REF:
n(Al2O3) =
p28
n(Al), n(O2) produced = 3 mol.
2. Which one of the following is a triprotic Lowry-BrØnsted acid?
A CH3COOH
B H3PO4
C H2CO3
D NH3
B
ANSWER:
FEEDBACK: In H3PO4, all three H atoms are bonded to a highly electronegative atom, and so can be
drawn away by a base. CH3COOH is a monoprotic acid, since only one H is bonded to
a highly electronegative atom. H2CO3 only has two H atoms that can be donated to a
base. NH3 is a weak base.
p22
REF:
3. The concentration of hydroxide ions in limewater, Ca(OH)2, of concentration 0.01 mol L-1 is:
A 10-12 mol L-1
B 0.005 mol L-1
C 0.01 mol L-1
D 0.02 mol L-1
D
ANSWER:
FEEDBACK: From the equation for the solution of Ca(OH)2, each 1 mol of Ca(OH)2 dissolving will
produce 2 mol OH- ions. Hence
[OH-] = 2 × [Ca(OH)2] = 2 × 0.01 mol L-1.
p25
REF:
4. The concentration of hydroxide ions in HCl of concentration 0.01 mol L-1 is:
A 10-12 mol L-1
B 0.01 mol L-1
C 2 mol L-1
D 12 mol L-1
A
ANSWER:
FEEDBACK: From the equation for the ionisation of HCl in water, each 1 mol of HCl reacting with
water will produce 1 mol H+ ions.
Hence [H+] = 0.01 mol L-1 = 10-2 mol L-1.
From the expression for Kw we have
REF:
[OH-] =
p24
= 10-12 mol L-1.
5. Which one of the following equations does NOT represent a Lowry-BrØnsted acid-base reaction?
A NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
B 2HCl(aq) + Mg(s)
MgCl2(aq) + H2(g)
C H2SO4(aq) + MgCO3(s)
MgSO4(aq) + H2O(l) + CO2(g)
D HSO4 (aq) + H2O(l)
SO42-(aq) + H3O+(aq)
ANSWER: B
0/1
POINTS:
FEEDBACK: In A, C and D, a proton transfer from one reactant to the other has occurred. (In the case
of C, the CO32- ions have accepted protons from the acid to form H2CO3, which has
broken down into H2O and CO2.) But in B, the oxidation number of Mg has increased
from 0 to +2, and of the H+ ions present has decreased from +1 to 0. Hence this is a
redox reaction, not an acid-base reaction.
pp21, 32
REF:
6. The pH of a solution of HNO3 of concentration 1 mol L-1 is:
A 0
B 1
C -1
D 14
A
ANSWER:
FEEDBACK: From the equation for the ionisation of HNO3 in water, each 1 mol of HNO3 reacting
with water will produce 1 mol H+ ions.
Hence [H+] = 1 mol L-1 = 100 mol L-1.
p24
REF:
7. Which statement about water in the following half-equation is CORRECT?
2H2O(l) + 2eH2(g) + 2OH-(aq)
A H2O has acted as a reductant only.
B H2O has acted as an oxidant only.
C H2O has acted both as an oxidant and a base.
D H2O has self-ionised.
B
ANSWER:
FEEDBACK: The half-equation shows that H2O has accepted electrons and hence has acted as an
oxidant in this reaction. This is supported by the fact that the oxidation number of the
key element, H, has decreased from +1 in H2O to 0 in H2. The H2O has not accepted
any protons, or it would form H3O+ ions, and hence also has not acted as a base. If it
had self-ionised, both H3O+ ions and OH- ions would be produced.
pp21, 23, 33
REF:
8. Which one of the following half-equations is correctly balanced?
A Cl2(g) + eCl-(aq)
B Al3+(aq)
Al(s) + 3e3+
C Fe (aq) + e
Fe2+(aq)
+
D NO3 (aq) + 2H (aq)
NO2(g) + H2O(l) + eANSWER:
C
FEEDBACK: In A, the number of Cl atoms is not balanced. In B and D, the charge is not balanced.
In C, the number of atoms of Fe and the charge are both balanced.
p35
REF:
9. The lead-based pigments in old paintings have often deteriorated into black lead(II) sulfide, PbS, due to
exposure to hydrogen sulfide in the air. To restore the original colour, the PbS is treated with hydrogen
peroxide solution, H2O2. The net equation for the reaction is:
PbS(s) + 4H2O2(aq)
PbSO4(s) + 4H2O(l)
In this reaction, in which one of the following capacities is the hydrogen peroxide acting?
A Acid
B Precipitating agent
C Reductant
D Oxidant
D
ANSWER:
FEEDBACK: In this case the key element in H2O2 is O. Since it has an oxidation number of -1 in
peroxides, its oxidation number decreases from -1 to -2 in the reaction. Therefore it has
been reduced and hence must be the oxidant.
p33
REF:
10. Which one of the following solutions would LOWER the pH if added to 50 mL of a certain solution of
pH of 4?
A 50 mL water
B 50 mL NaOH solution of concentration 4 mol L-1
C 50 mL HCl of concentration 0.0001 mol L-1
D 50 mL HCl of concentration 4 mol L-1
D
ANSWER:
FEEDBACK: The addition of water will raise the pH of an acid solution so that it is closer to 7, the
pH of pure water. The addition of the base NaOH also will decrease [H+] and hence
increase the pH. The solution in C has [H+] = 10-4 mol L-1 and so its pH is also 4. Thus
the pH will not change when it is added. Since the acid in D has a much higher
concentration, its [H+] will be much higher than that of the original solution and its
addition will lower the pH.
p24
REF:
11. One substance that can cause acid rain is nitrogen dioxide gas, NO2. What is the mass, in grams, of one
molecule of nitrogen dioxide?
A
B
C 46.0 × 6.02 × 1023
D 46.0
A
ANSWER:
FEEDBACK: We can apply the formulas relating the amount of substance, n, with the number of
particles and with mass, or we can reason this out as follows. The molar mass of NO2
is 46.0 g mol-1, which is the total mass of
6.02 × 1023 molecules. Hence the mass of one molecule is obtained by dividing the
total mass of
6.02 × 1023 molecules by
6.02 × 1023.
p4
REF:
12. If 50.0 mL of HCl of concentration 0.300 mol L-1 is added to 50.0 mL of NaOH of concentration 0.100
mol L-1, what will be the pH of the final solution?
A 0.100
B 1.00
C 7.00
D 13.0
B
ANSWER:
FEEDBACK: n(HCl) added = 0.0150 mol and n(NaOH) present = 0.00500 mol. From the equation
for the reaction, n(HCl) required to react with the NaOH present = 0.00500 mol, so
HCl is in excess by (0.0150 – 0.00500) mol = 0.0100 mol.
Hence [HCl] in 100.0 mL final solution
REF:
=
= 0.100 mol L-1.
From this we deduce
[H+] = 10-1.00 and the pH = 1.00.
pp28–9
13. Ozone in the air at ground level presents a serious pollution problem. A 1.0 m3 sample of air collected
at 100 kPa and 24 °C was found to contain 38.5 mL of ozone. The amount of ozone gas in the sample
would be closest to:
A 0.001 56 mol
B 0.0193 mol
C 1.56 mol
D 19.3 mol
A
ANSWER:
FEEDBACK: We substitute into PV = nRT, where P = 100 kPa,
V = 38.5 × 10-3 L and T = 297 K. The other responses result from failing to perform
one or both of the required unit conversions.
p7
REF:
14. The formula NOx represents the family of oxides of nitrogen. One oxide was found to contain 30.4 %
nitrogen by mass. What was the empirical formula of the oxide?
A N2O
B N2O4
C NO2
D NO
C
ANSWER:
FEEDBACK: Both B and C contain 30.4 % nitrogen by mass, but only NO2 is classified as an
empirical formula – that is, a formula showing the simplest ratio of atoms.
pp15–16
REF:
15. To deduce the molecular formula of a certain hydrocarbon, an experiment was conducted in which
lithium hydroxide was used to absorb carbon dioxide produced in the combustion of the hydrocarbon.
Lithium hydroxide reacts with the carbon dioxide according to this equation:
LiOH(s) + CO2(g)
LiHCO3(aq)
What is the maximum volume of CO2 that can be absorbed at SLC by 10 g of lithium hydroxide?
A 10 mL
B 98 mL
C 9.8 L
D 10 L
ANSWER:
D
FEEDBACK: From the equation, n(CO2) absorbed = n(LiOH), hence the maximum amount of CO2
absorbed will be the amount that completely reacts with 10 g of LiOH, which is 0.42
mol LiOH. From the above mole relationship, this means that 0.42 mol CO2 can be
REF:
absorbed. Since the gas is measured at SLC, we simply substitute into n =
the volume of CO2 absorbed.
pp5, 12–13
to find
16. If 0.5 mol of C6H14 is burnt completely in air at 1000 °C and 760 mmHg, what will be the total volume
of the gas products?
A 20 L
B 300 L
C 600 L
D 700 L
D
ANSWER:
FEEDBACK: From the balanced equation for the combustion, 1 mol C6H14 produces 6 mol CO2 and
7 mol H2O, which is 13 mol gas products altogether, since at this T, H2O is a gas. Thus
0.5 mol C6H14 produces 6.5 mol gas products. Substituting n = 6.5 mol, p = 101.325
kPa and T = 1273 K into pV = nRT gives V = 700 L (to 1 significant figure).
p7
REF:
17. Two gas syringes each contain the same volume of gas at 25 °C. One of these gas syringes contains
0.010 g neon gas at 100 kPa, and the other contains 0.010 g propane gas, C3H8. The pressure exerted by
the propane gas on its gas syringe will be:
A 46 kPa
B 100 kPa
C 200 kPa
D 220 kPa
A
ANSWER:
FEEDBACK: From pV = nRT we see that if two samples of gas occupy the same volume at the same
T, the pressure they exert on their container is directly proportional to their amount, in
mol.
Hence we have:
.
Subsituting into this equation, and transforming and cancelling gives:
= 46 kPa.
REF:
p7
18. Ammonia gas can react with oxygen gas to produce nitrogen gas and steam, according to this equation:
4NH3(g) + 3O2(g)
2N2(g) + 6H2O(g)
If 2.0 L of ammonia is mixed with a certain volume of excess oxygen and the final volume is 6.0 L,
with all gases measured at the same temperature and pressure, the volume of oxygen originally added
must be:
A 1.5 L
B 2.0 L
C 3.5 L
D 4.0 L
C
ANSWER:
FEEDBACK: From the equation, 4 mol NH3 reacting produces 8 mol gas products. Hence n(gas
products) = 2n(NH3) and therefore V(gas products) = 2V(NH3), since the gases are at
the same T, P. Thus if 2.0 L NH3 reacts, 4.0 L gas products are formed, which means
(6.0 – 4.0) L of O2 remains unreacted.
REF:
Now V(O2) reacting = V(NH3) = 1.5 L,
so V(O2) originally added = (1.5 + 2.0) L = 3.5 L.
p10
19. 1 L of a certain hydrocarbon was burnt in excess air at 500 °C and 1 atm pressure. The carbon dioxide
and steam produced in this reaction were found to occupy 5 L and 6 L respectively when measured at
500 °C and 1 atm pressure. What was the molecular formula of the hydrocarbon?
A C3H8
B C5H6
C C5H12
D C6H12
C
ANSWER:
FEEDBACK: When gases are compared at the same T, P, their amount in moles is directly
proportional to their volume. Hence the equation for the combustion must be of the
form:
CxHy + pO2
5CO2 + 6H2O
(States are omitted for simplicity.) For this to balance,
x = 5, y = 12 (and p = 8).
pp13–14
REF:
20. Acidified potassium dichromate, K2Cr2O7, reacts with ethanol, C2H5OH, according to the following
ionic equation:
2Cr2O72-(aq) + 16H+(aq) + 3C2H5OH(aq)
4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)
A 20.00 mL aliquot of a standard solution of acidified potassium dichromate, of concentration 0.07500
mol L-1, was reacted with a solution of ethanol. The mean volume of the ethanol solution required for
the titration was 25.32 mL. The concentration of ethanol in the solution, in mol L-1, is:
A 0.03949
B 0.05917
C 0.08886
D 0.1185
ANSWER: C
FEEDBACK: Since there is one Cr2O72- ion per formula unit,
[Cr2O72-] = [K2Cr2O7]. Using n = cV, we have
n(Cr2O72-) = 0.001 500 mol.
From the equation,
REF:
n(C2H5OH) = × n(Cr2O72-) = 0.002250 mol. Dividing by V(C2H5OH) we get
[C2H5OH] = 0.08886 mol L-1.
p9