Chemistry 11 (HL)
Unit 3 / IB Topics 1.3-1.5
Unit 3 Review: Quantities in Chemical Reactions
Paper 2 SOLUTIONS
PAPER 2 STYLE QUESTIONS
14.
0.502 g of an alkali metal sulfate is dissolved in water and excess barium chloride solution,
BaCl2(aq) is added to precipitate all the sulfate ions as barium sulfate, BaSO4(s). The
precipitate is filtered and dried and weighs 0.672 g.
(a)
Calculate the amount (in mol) of barium sulfate formed. (2)
mol BaSO 4 = 0.672 g x
(b)
1 mol
= 0.00288 mol
233.39g
Determine the amount (in mol) of the alkali metal sulfate present.
(1)
The alkali metal sulfate is an alkali metal (Group 1) ion bonded to a sulfate ion. It
has a general formula of X2SO4.
mol X 2SO 4 = mol BaSO 4 = 0.00288 mol
(c)
Determine the molar mass of the alkali metal sulfate and state its units. (2)
molar mass X 2SO 4 =
(d)
mass X 2SO 4
0.502 g
=
= 174.31 g mol−1
moles X 2SO
0.00288 mol
Deduce the identity of the alkali metal, showing your workings. (2)
molar mass X2SO4 = 2(molar mass X) + molar mass S + 4(molar mass O)
174.31 = 2x + 32.06 + 4(16.00)
x = 39.13
The alkali metal (Group 1) with a molar mass of 39.13 is potassium (K).
(e)
Write an equation for the precipitation reaction, including state symbols. (2)
K2SO4(aq) + BaCl2(aq)  BaSO4(s) + 2 KCl(aq)
(Total 9 marks)
p. 1
Chemistry 11 (HL)
15.
Unit 3 / IB Topics 1.3-1.5
State and explain what would happen to the pressure of a given mass of gas when its absolute
temperature and volume are both doubled. (Total 3 marks)
Doubling the absolute (Kelvin) temperature will double the pressure.
Doubling the volume will reduce the pressure by ½.
Therefore the net effect is NO CHANGE in pressure.
You could show this mathematically:
P1V1
PV
= 2 2
T1
T2
P1V1
P (1/2 V1 )
= 2
T1
2T1
P2 =
P1 V1 (2T1 )
(1/ 2 V1 ) T1
P2 = P1
p. 2
Chemistry 11 (HL)
16.
Unit 3 / IB Topics 1.3-1.5
A toxic gas, A, consists of 53.8% nitrogen and 46.2% carbon by mass. At 273 K and
5
3
1.01×10 Pa, 1.048 g of A occupies 462 cm . Determine the empirical formula of A.
Calculate the molar mass of the compound.
(Total 2 marks)
STEP 1: Find the empirical formula of gas A using the % composition data.
Assume the sample of gas A has 53.8 g of nitrogen and 46.2 g of carbon.
mol N = 53.8 g x
1 mol
= 3.84 mol
14.01g
mol C = 46.2 g x
1 mol
= 3.84 mol
12.01g
lowest mole ratio of N : C is 1 : 1
Therefore the empirical formula is NC (or CN)
STEP 2: Find the molar mass of gas A using the Ideal Gas Law:
APPROACH A – two steps:
PV = nRT
(101 kPa) (0.462 dm3 ) = n (8.31 J mol−1 K −1 ) (273 K)
n = 0.0206 mol
m
n
1.048 g
=
0.0206 mol
= 50.9 g mol−1
molar mass =
APPROACH B – one step using the alternate form of the Ideal Gas Law:
⎛ m⎞
PV = ⎜ ⎟ RT
⎝ M⎠
⎛ 1.048 g ⎞
(101 kPa) (0.462 dm3 ) = ⎜
(8.31 J mol−1 K −1 ) (273 K)
⎟
⎝ M ⎠
M = 50.9 g mol−1
p. 3
Chemistry 11 (HL)
17.
Unit 3 / IB Topics 1.3-1.5
The reaction below represents the reduction of iron ore to produce iron.
2Fe2O3 + 3C  4Fe + 3CO2
A mixture of 30 kg of Fe2O3 and 5.0 kg of C was heated until no further reaction occurred.
Calculate the maximum mass of iron that can be obtained from these masses of reactants.
(Total 5 marks)
STEP 1: Find the moles of each reactant:
mol Fe2O3 = 3.0 x 104 g x
mol C = 5.0 x 103 g x
1 mol
= 1.87 x 102 mol
159.70 g
1 mol
= 4.16 x 102 mol
12.01 g
STEP 2: Find the limiting reactant:
APPROACH A: Find the moles of iron formed if each reactant is used up:
If all Fe2O3 is used up:
mol Fe = 1.87 x 102 mol Fe2O3 x
4 mol Fe
= 3.74 x 102 mol
2 mol Fe2O3
If all C is used up:
mol Fe = 4.16 x 102 mol C x
4 mol Fe
= 5.54 x 102 mol
3 mol C
The limiting reactant is Fe2O3 because it produces less product (Fe).
APPROACH B: Divide the moles of each reactant by the co-efficient in the equation:
Fe2O3 :
1.87 x102 mol
= 93.5
2
The limiting reactant is Fe2O3.
2
C:
4.16 x 10 mol
= 208
3
STEP 3: Find the moles of Fe formed using the limiting reactant (Fe2O3) and the mole ratio:
mol Fe = 1.87 x 102 mol Fe2O3 x
4 mol Fe
= 3.74 x 102 mol
2 mol Fe2O3
STEP 4: Find the mass of Fe:
mass Fe = 3.74 x 102 mol x
55.85 g
= 2.1 x 104 g = 21 kg
1 mol
p. 4
Chemistry 11 (HL)
18.
(a)
Unit 3 / IB Topics 1.3-1.5
Write an equation for the formation of zinc iodide from zinc and iodine.
(1)
Zn + I2  ZnI2
(b)
100.0 g of zinc is allowed to react with 100.0 g of iodine producing zinc iodide.
Calculate the amount (in moles) of zinc and iodine, and hence determine which reactant
is in excess. (3)
STEP 1: Find the moles of each reactant:
mol Zn = 100.0 g x
1 mol
= 1.530 mol
65.38 g
mol I2 = 100.0 g x
1 mol
= 0.3940 mol
253.80 g
STEP 2: Find the excess reactant (read the question carefully):
Since the Zn : I2 mole ratio is 1 : 1, it is clear that Zn is the excess reactant.
( 1.530 mol Zn > 0.394 mol I2 )
(c)
Calculate the mass of zinc iodide that will be produced.
(1)
STEP 1: Find the moles of ZnI2 formed using the limiting reactant (I2) and the mole
ratio:
mol ZnI2 = 0.3940 mol I2 x
1 mol ZnI2
= 0.3940 mol
1 mol I2
STEP 2: Find the mass of ZnI2:
mass ZnI2 = 0.3940 mol x
319.18 g
= 125.8 g
1 mol
p. 5
Chemistry 11 (HL)
19.
Unit 3 / IB Topics 1.3-1.5
Potassium hydroxide and calcium chloride solutions are mixed together, and a hydroxide precipitate
forms.
a)
Write the complete balanced chemical equation for this reaction.
2 KOH(aq) + CaCl2(aq)  Ca(OH)2(s) + 2 KCl(aq)
b)
If you mix 45.0 mL of 0.85 M potassium hydroxide with excess calcium chloride,
what mass of precipitate should form?
STEP 1: Find the moles of KOH using C = n/V:
mol KOH = C x V
= 0.85 M x 0.045 L
= 0.038 mol
STEP 2: Find the moles of precipitate (Ca(OH)2) using the mole ratio:
mol Ca(OH)2 = 0.038 mol KOH x
1 mol Ca(OH)2
= 0.019 mol
2 mol KOH
STEP 3: Find the mass of Ca(OH)2
mass Ca(OH)2 = 0.019 mol x
74.10 g
= 1.4 g
1 mol
p. 6
Chemistry 11 (HL)
20.
Unit 3 / IB Topics 1.3-1.5
Freon-12, CCl2F2, is prepared from CCl4 by reaction with HF. HCl is the other product of this reaction.
What is the percent yield of a reaction in which 12.5 g of Freon-12 is produced from 32.9 g of carbon
tetrachloride?
Write the chemical equation first:
CCl4 + 2 HF  CCl2F2 + 2 HCl
STEP 1: Find the moles of CCl4 used:
mol CCl4 = 32.9 g x
1 mol
= 0.214 mol
153.81 g
STEP 2: Find the theoretical moles of CCl2F2 produced:
mol CCl2F2 = 0.214 mol CCl4 x
1 mol CCl2F2
= 0.214 mol
1 mol CCl4
STEP 3: Find the theoretical mass of CCl2F2 produced:
mass CCl2F2 = 0.214 mol x
120.91 g
= 25.9 g
1 mol
STEP 4: Find the percentage yield:
actual yield
x 100
theoretial yield
12.5 g
=
x 100
25.9 g
= 48.3%
percentage yield =
p. 7
Chemistry 11 (HL)
21.
Unit 3 / IB Topics 1.3-1.5
The mass of H2 produced by reaction of 1.80 g Al and 6.00 g of sulfuric acid is 0.112 g. What is the
percent yield?
2 Al + 3 H2SO4 ⇒ Al2(SO4)3 + 3 H2
ALWAYS BALANCE THE EQUATION FIRST!
STEP 1: Find the moles of each reactant:
mol Al = 1.80 g x
1 mol
= 0.0667 mol
26.98 g
mol H2SO 4 = 6.00 g x
1 mol
= 0.0612 mol
98.08 g
STEP 2: Find the limiting reactant:
APPROACH A: Find the moles of product (H2) formed if each reactant is used up:
If all Al is used up:
mol H2 = 0.0667 mol Al x
3 mol H2
= 0.100 mol
2 mol Al
If all H2SO4 is used up:
mol H2 = 0.0612 mol H2SO 4 x
3 mol H2
= 0.0612 mol
3 mol H2SO 4
The limiting reactant is H2SO4 because it produces less product (H2).
APPROACH B: Divide the moles of each reactant by the co-efficient in the equation:
Al:
0.0667 mol
= 0.0334
2
The limiting reactant is H2SO4.
H2SO 4 :
0.0612 mol
= 0.0204
3
STEP 3: Find the theoretical moles of H2 formed using the limiting reactant (H2SO4) and the
mole ratio:
mol H2 = 0.0612 mol H2SO 4 x
3 mol H2
= 0.0612 mol
3 mol H2SO 4
STEP 4: Find the theoretical mass of H2:
mass H2 = 0.0612 mol x
2.02 g
= 0.124 g
1 mol
STEP 5: Find the percentage yield:
percentage yield =
=
actual yield
x 100
theoretial yield
0.112 g
x 100
0.124 g
= 90.3%
p. 8
Chemistry 11 (HL)
22.
Unit 3 / IB Topics 1.3-1.5
An experiment requires a 0.125 mol/L solution of magnesium chloride. The stock bottle of magnesium
chloride only contains 42.5 g of the solid. What is the maximum volume of solution that can be
prepared?
STEP 1: Find moles of solid MgCl2 available:
mol MgCl2 = 42.5 g x
1 mol
= 0.446 mol
95.21 g
STEP 2: Find the volume of solution that could be prepared using C = n/V:
n
V
0.446 mol
0.125 mol/L =
V
V = 3.57 L
C=
23.
What is the new concentration of a solution if 300.0 mL of water is added to 100.0 mL of a
0.025 mol/L CaCl2 solution?
Use the dilution formula!
C1V1 = C2 V2
(0.025 mol/L) (100.0 mL) = C2 (100.0 + 300.0 mL)
C2 = 0.00625 mol/L
p. 9
Chemistry 11 (HL)
Unit 3 / IB Topics 1.3-1.5
24. What is the density of ammonia gas, in g/L, at STP?
Recall: density = mass / volume.
Assume the gas has a volume of 1 L.
Use the Ideal Gas Law to find the mass of ammonia with a volume of 1 L at STP:
⎛ m⎞
PV = ⎜ ⎟ RT
⎝ M⎠
⎛
⎞
m
(101 kPa) (1 L) = ⎜
(8.31 J mol−1 K −1 ) (273 K)
−1 ⎟
⎝ 17.04 g mol ⎠
m = 0.758 g
Now find the density:
D=
25.
m
0.758 g
=
= 0.758 g/L
V
1L
What is the volume occupied by 0.38 mol of sulfur trioxide gas at STP? Show 2 different ways of
finding this value.
METHOD 1: Use the molar volume of a gas at STP. (Remember all gases at STP have a molar
3
volume of 22.4 dm per mole. This value is in Table 2 of the Data Booklet.)
V = 0.38 mol x
22.4 dm3
1 mol
= 8.5 dm3
5
METHOD 2: Use the Ideal Gas Law. Remember that STP conditions are a pressure of 1.01 x 10 Pa
(101 kPa) and a temperature of 273 K.
PV = nRT
(101 kPa) V = (0.38 mol) (8.31 J mol−1 K −1 ) (273 K)
V = 8.5 dm3
p. 10
Chemistry 11 (HL)
26.
Unit 3 / IB Topics 1.3-1.5
3
-3
3
-3
50.00 cm of 0.400 mol dm CaCl2 is mixed with 25.00 cm of 0.900 mol dm AgNO3. A precipitate of
silver chloride, AgCl, forms and it is collected by filtration. The mass of the filter paper before filtration
was 1.250 g. After filtration and drying, the mass of the filter paper was 4.205 g.
Write the balanced equation for the reaction first:
CaCl2(aq) + 2 AgNO3(aq)  2 AgCl(s) + 2 NaNO3(aq)
a)
Determine the limiting reactant.
STEP 1: Find the moles of each reactant, using C = n/V:
mol CaCl2 = C x V
= 0.400 mol dm−3 x 0.05000 dm3
= 0.0200 mol
mol AgNO3 = C x V
= 0.900 mol dm−3 x 0.02500 dm3
= 0.0225 mol
STEP 2: Determine the limiting reactant:
METHOD A: Find the moles of AgCl produced if each reactant was used up:
If all CaCl2 is used up:
mol AgCl = 0.0200 mol CaCl2 x
2 mol AgCl
= 0.0400 mol
1 mol CaCl2
If all AgNO3 is used up:
mol AgCl = 0.0225 mol AgNO3 x
2 mol AgCl
= 0.0225 mol
2 mol AgNO3
The limiting reactant is AgNO3 because it forms the lower amount of product (AgCl).
METHOD B: Divide the moles of each reactant by its co-efficient from the equation:
CaCl2 :
AgNO3 :
0.0200 mol
= 0.0200
1
0.0225 mol
= 0.0113
2
The limiting reactant is AgNO3
p. 11
Chemistry 11 (HL)
b)
Unit 3 / IB Topics 1.3-1.5
Calculate the theoretical yield of the precipitate that forms.
STEP 1: Find the number of moles of precipitate (AgCl), using the limiting reactant and
mole ratio:
mol AgCl = 0.0225 mol AgNO3 x
2 mol AgCl
= 0.0225 mol
2 mol AgNO3
STEP 2: Find the mass of precipitate (AgCl):
mass AgCl = 0.0225 mol x
c)
143.32 g
= 3.22 g
1 mol
Calculate the actual yield.
actual yield = mass of precipitate formed in the experiment
= (mass of filter paper with precipitate) – (mass of filter paper only)
= 4.205 g – 1.250 g
= 2.955 g
d)
Calculate the percentage yield.
actual yield
x 100
theoretial yield
2.955 g
=
x 100
3.22 g
= 91.8%
percentage yield =
e)
Calculate the mass of excess reactant that remains.
Excess reactant = CaCl2
STEP 1: Find moles of CaCl2 used in the reaction:
mol CaCl2 = 0.0225 mol AgNO3 x
1 mol CaCl2
= 0.0113 mol
2 mol AgNO3
STEP 2: Find moles of CaCl2 remaining:
mol CaCl2 remaining = (initial mol CaCl2 ) - (mol CaCl2 used)
= 0.0200 mol - 0.0113 mol
= 0.0087 mol
STEP 3: Find mass of CaCl2 remaining:
mass CaCl2 = 0.0087 mol x
110.98 g
= 0.97 g
1 mol
p. 12