DATE: NAME: CLASS:
CHAPTER 8
Limiting Reactant Problems
BLM 8.1.2ASSESSMENT
1. Extraction of zinc from zinc oxide takes place as follows: ZnO(s) + C(s) → Zn(s) + CO(g)
In an industrial setting, 17.2 mol of zinc oxide are reacted with 43.2 mol of carbon in the form of charcoal. Identify the limiting reactant.
2. If 1.00 kg of glucose (5.55 mol) is reacted with 34.0 mol oxygen during the process of cellular respiration, which reactant is limiting? Why does your answer seem logical?
3. Acrylic, a common synthetic fibre, is formed from acrylonitrile. Acrylonitrile is formed in the following reaction:
4C3H6(g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g)
What is the limiting reactant when 126 g of C3H6(g) reacts with 175 g of NO?
CHAPTER 8
Limiting Reactant Problems
(continued)BLM 8.1.2
ASSESSMENT
5. One method of ammonia production occurs when lithium nitride reacts with water to produce ammonia and lithium hydroxide. If 4.87 g of lithium nitride reacts with 7.74 g of water, find the limiting reactant.
6. When 0.25 g of aluminium reacts with 0.51 g of copper(II) chloride, which reactant is limiting?
7. A 33.76 g mass of zinc reacts with 54.08 g of hydrogen chloride. Which reactant is in excess?
8. Chloride dioxide is a reactive oxidizing agent. It is used to purify water. The products of the reaction are chloric acid and hydrochloric acid. If 71.00 g of chloride dioxide is mixed with 19.00 g of water, what is the limiting reactant?
Limiting Reactant Problems
Answer Key
ANSWER KEY
1. ZnO(s) + C(s) → Zn(s) + CO(g)
Zinc oxide is the limiting reactant.
2. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)
Glucose is the limiting reactant. This seems reasonable because oxygen is quite often in abundance.
3. 4C3H6(g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g)
C3H6 is the limiting reactant.
4. CaF2(s) + H2SO4(aq) → CaSO4(s) + 2HF(g)
5. Li3N(s) + 3H2O(ℓ) → 3LiOH(aq) + NH3(g)
Lithium nitride is the limiting reactant.
CHAPTER 8
Limiting Reactant Problems
Answer Key
(continued)BLM 8.1.2A
ANSWER KEY
6. 2Al(s) + 3CuCl2(aq) → 3Cu(s) + 2AlCl3(aq)
Copper(II) chloride is limiting.
7. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Hydrogen chloride is in excess. (Zinc is limiting.)
8. ClO2(g) + H2O(ℓ) → HCl(aq) + HClO3(aq)
DATE: NAME: CLASS:
CHAPTER 8
Expected Quantity of Product
Problems
BLM 8.1.3
ASSESSMENT
1. Manganese(III) fluoride is formed in the following reaction: 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ)
1.24 g of manganese(II) iodide reacts with 25.0 g of fluorine gas. What mass of manganese(III) fluoride is expected?
2. Copper metal reacts with nitric acid according to the following reaction: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ)
What mass of NO is produced when 57.4 g of Cu reacts with 165 g of HNO3?
3. Hydrogen peroxide, H2O2, decomposes to liquid water and oxygen gas. Calculate the mass of
CHAPTER 8
Expected Quantity of Product
Problems
(continued)BLM 8.1.3
ASSESSMENT
4. When ammonia reacts with oxygen gas, the products are nitrogen dioxide gas and water vapour. A 1.30 kg mass of ammonia is present in 4.21 kg of oxygen gas. What mass of nitrogen dioxide is expected?
CHAPTER 8
Expected Quantity of Product
Problems Answer Key
BLM 8.1.3A
ANSWER KEY
1. 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ)
MnI2(s) is the limiting reactant, and the expected mass of manganese(III) fluoride is 0.450g. 2. 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ)
Copper is the limiting reactant, and the expected mass of NO(g) is 18.1g.
3. 2H2O2(ℓ) → 2H2O(ℓ) + O2(g)
(ℓ)
The expected mass of water is 5.30 g.
4. 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
The expected mass of nitrogen dioxide is 3.46 kg. Oxygen gas is limiting.
5. Fe2(SO4)3(s) + 6NaOH(aq) → 2Fe(OH)3 (aq)+ 3Na2SO4(aq)
CHAPTER 8
Process for Solving Limited
Reactant Solution Stoichiometry
Problems
BLM 8.1.4
OVERHEAD
1.
Write the complete balanced equation for the reaction.
2.
Determine the molar amount of each reactant,
n
reactant, using the
equation
n
=
V
×
c
, where
c
is concentration (mol/L) and
V
is volume (L).
3.
Determine the molar amount of the product formed using the mole
ratio
based on the balanced equation and
n
reactantfor EACH reactant. The
equation for this calculation is
n
product=
n
reactant× mole ratio.
4.
The masses of reactants and product can also be calculated using
the equation
m
=
n
×
M
, with the appropriate
n
and
M
values for
each reactant and product.
5.
The reactant that results in the production of the LEAST amount of
product is the limiting reactant. It will limit the amount of product
that is formed.
CHAPTER 8
Limiting Reactant in Solution and
Gas Stoichiometry
BLM 8.1.5
ASSESSMENT
1. (a) When 50.0 mL of 0.100 mol/L silver nitrate reacts with 25.0 mL of 0.150 mol/L sodium chromate, which reactant is limiting?
AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + NaNO3(aq)
(b) What mass of precipitate is formed?
2. A quantity of sodium sulfide with a mass of 8.76 g is added to 350 mL of 0.250 mol/L lead(II) nitrate solution. Which reactant is limiting?
3. A 25.0 mL volume of 0.400 mol/L lead(II) nitrate is mixed with 300 mL of 0.220 mol/L potassium iodide. What is the maximum mass of precipitate that can form?
Limiting Reactant in Solution and
Gas Stoichiometry
(continued)ASSESSMENT
4. A 3.7 L volume of propane gas combusts in the presence of 11.2 L of oxygen gas, both at STP. Which reactant is limiting?
CHAPTER 8
Limiting Reactant in Solution and
Gas Stoichiometry Answer Key
BLM 8.1.5A
ANSWER KEY
1. (a) 2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq)
Silver nitrate is limiting.
(b)
2. Na2S(s) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbS(s)
Lead(II) nitrate is limiting.
3. Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Lead(II) nitrate is limiting.
CHAPTER 8
Limiting Reactant in Solution and
Gas Stoichiometry Answer Key
(continued)
BLM 8.1.5A
ANSWER KEY
4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Oxygen is the limiting reactant.
5. H2(g) + O2(g) → H2O(ℓ)
The quantity of liquid water expected is 1.50 g.
CHAPTER 8
Percentage Yield Problems
BLM 8.2.1ASSESSMENT
1. N2(g)+ 3H2(g) → 2 NH3(g)
When 7.5 × 101 g of nitrogen gas reacts with sufficient hydrogen gas, the theoretical yield of
ammonia is 9.10g. If 1.72 g of ammonia is obtained by experiment, what is the percentage yield of the reaction?
2. 20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
HBrO3(aq) + 5HBr(aq) → 3H2O(ℓ) + 3Br2(aq) (a) What is the predicted yield of Br2 for this reaction?
(b)If 47.3 g of Br2 is produced, what is the percentage yield of Br2?
3. In order to produce a lead(II) chromate precipitate, lead(II) chloride reacts with sodium chromate in solution. A 12.5 g mass of lead(II) chloride is mixed into solution, and is allowed to react with excess sodium chromate.
(a) What is the predicted yield of lead(II) chromate?
Percentage Yield Problems
(continued)ASSESSMENT
(b)Calculate the percentage yield if 13.8 g of lead(II) chromate is produced experimentally.
4. When calcium carbonate reacts with hydrogen chloride, the products are calcium chloride, carbon dioxide and water. If this reaction occurs with 81.5% yield, what mass of carbon dioxide will be collected if 15.7 g of calcium carbonate is added to sufficient hydrogen chloride?
Percentage Yield Problems
Answer Key
ANSWER KEY
1.
2. HBrO3(aq) + 5HBr(aq) → 3H2O(ℓ) + 3Br2(aq)
(a)
The predicted yield of Br2 is 74.4 g.
(b)
3. PbCl2(s) + Na2CrO4(aq) → PbCrO4(s) + 2NaCl(aq)
(a)
The predicted yield of lead(II) chromate is 14.5 g.
CHAPTER 8
Percentage Yield Problems Answer
Key
(continued)BLM 8.2.1A
ANSWER KEY
4. CaCO3(s) + 2HCl(aq) → CaCl2(s) + CO2(g) + H2O(ℓ)
CHAPTER 8
Acid-Base Titration Answer Key
BLM 8.3.1AANSWER KEY
A titration setup
Additional Titration Terminology
Equivalence point stage of titration at which stoichiometrically equivalent amounts of reactant have been consumed
Endpoint the point at which the indicator changes colour
Standardization determination of the concentration of a solution by analyzing its reaction with a solution of known concentration (a standard)
pH the negative logarithm (base-ten) of the molar concentration of hydronium ions in a solution
Indicator a substance that changes colour over a given pH range; usually a weak, monoprotic acid
Titration Curve a plot of the pH of a reaction mixture versus the volume of titrant added burette – a graduated glass
tube with a tap at one end; used to measure the volume of liquid during a titration
titrant – solution that is being added and having its volume measured during a titration
Erlenmeyer flask – a cone-shaped glass container with a narrow neck and a flat, circular base; contains the solution to be titrated
burette clamp – a metal apparatus that mounts on the retort stand to hold the burette in place
retort stand – a narrow rod supported vertically on a flat, rectangular base; used to hold the burette upright above the Erlenmeyer flask
CHAPTER 8
Titration Step-By-Step
BLM 8.3.2OVERHEAD
1.
A sample to be analyzed is measured in a
graduated or volumetric pipette and delivered to
an Erlenmeyer flask.
2.
A few drops of the appropriate indicator are
added to the sample being analyzed.
3.
After taking an initial reading of the volume of titrant in the burette,
the titrant is added, slowly, to the sample in the Erlenmeyer flask,
while the sample is swirled.
4.
The
endpoint of
the titration occurs when the indicator colour changes dramatically
and permanently. With a well-chosen indicator, the difference in the
volumes of titrant required to reach the endpoint and the equivalence
point should differ only by one drop of titrant (approximately 0.05
mL).
CHAPTER 8
Titration Problems
BLM 8.3.3ASSESSMENT
1. (a) Use a diagram to illustrate the neutralization of sulfuric acid with sodium hydroxide. Use H+
and OH– in your illustration.
(b) According to the diagram, how many moles of sodium hydroxide must be present to neutralize 1 mol of sulfuric acid?
(c) How many moles of phosphoric acid are required to neutralize 7.2 mol of potassium hydroxide?
2. A 17.85 mL volume of nitric acid neutralizes 25.00 mL of 0.150 mol/L sodium hydroxide. What is the concentration of the nitric acid when the pH is exactly 7.0?
3. A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate is formed. What is the concentration of the sulfuric acid?
4. The following data were collected during a titration. Calculate the concentration of the sodium hydroxide solution.
Volume of HCl(aq) 10.00 mL
Initial volume of NaOH(aq) 23.08 mL
Final volume of NaOH(aq) 1.06 mL
Concentration of HCl(aq) 0.235 mol/L
Titration Problems
(continued)ASSESSMENT
5. What volume of 0.250 mol/L sulfuric acid is needed to react completely with 37.2 mL of 0.650 mol/L potassium hydroxide?
6. It is found that 42.5 mL of 1.02 mol/L NaOH have been added to 50.0 mL of vinegar (ethanoic acid) when the phenolphthalein in the solution just turns pink. What is the concentration of the vinegar?
7. A student conducts three trials to determine the concentration of barium hydroxide. The titrant used is a 0.250 mol/L hydrochloric acid solution. Each sample of barium hydroxide is 10.00 mL. Calculate the concentration of the barium hydroxide for each trial. What is the average concentration of barium hydroxide?
Trial # 1 2 3
Final volume HCl(aq) (mL) 37.32 24.56 11.78
CHAPTER 8
Titration Problems Answer Key
BLM 8.3.3AANSWER KEY
1. (a)
(b) According to the diagram, two moles of sodium hydroxide must be present to neutralize one mole of sulfuric acid.
(c) H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(ℓ)
1.4 moles of phosphoric acid are required to neutralize 7.2 moles of potassium hydroxide.
2. HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(ℓ)
When the pH is 7.0, the concentration of the nitric acid is 0.210 mol/L.
3. H2SO4(aq) + 2NH3(aq) → (NH4)2SO4(aq)
The concentration of the sulfuric acid is 0.0761 mol/L.
4. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ)
H
+H
+H
+SO
42–
+
Na
+Na
OH
+ –OH
–Na
+SO
42–Na
++
OH
–OH
–The concentration of the sodium hydroxide solution was 0.107mol/L.
CHAPTER 8
Titration Problems Answer Key
(continued)
BLM 8.3.3A
ANSWER KEY
5. H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(ℓ)
48.4 mL of sulfuric acid is needed to react completely.
6. CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(ℓ)
The concentration of ethanoic acid is 0.867 mol/L.
CHAPTER 8
pH Titration Curves and Indicators
BLM 8.3.5OVERHEAD
The graph on the left shows the pH curve obtained when a strong base is
added to a strong acid. The reaction mixture starts off very acidic and
becomes very basic. The opposite is true when a strong acid is added to a
strong base, as shown in the graph on the left.
Indicators Suitable for Strong Acid-Strong Base Titrations
Indicator
pH range
Colour change as pH increases
bromocresol green
3.8–5.4
yellow to blue
methyl red
4.8–6.0
red to yellow
chlorophenol red
5.2–6.8
yellow to red
bromothymol blue
6.0–7.6
yellow to blue
phenol red
6.6–8.0
yellow to red
phenolphthalein
8.2–10.0
colourless to pink
DATE: NAME: CLASS:
Selecting an Indicator
ASSESSMENT
1. A titration is performed and the following pH curve is generated.
(a) Select an indicator that would be appropriate to use in this titration.
(b) What is the endpoint in this titration, given the indicator selected in part (a)?
(c) What is the equivalence point in this titration?
(d) What colour change would occur?
2. A titration is performed and the following pH curve is generated.
(a) Select an indicator that would be appropriate to use in this titration.
(b) What is the endpoint in this titration, given the indicator selected in part (a)?
(c) What is the equivalence point in this titration?
(d) What colour change would occur?
CHAPTER 8
Selecting an Indicator Answer
Key
BLM 8.3.6A
ANSWER KEY
2.
(e) Phenol red is an appropriate indicator to use in this titration, since it changes colour between pH 6.6 and pH 8.
(f) The endpoint in this titration would be approximately 6.6 if phenol red were used as an indicator.
(g) The equivalence point in this titration is approximately pH 7.
(h) The indicator would change from yellow to red.
2.
(a) Chlorophenol red is an
appropriate indicator to use in this titration, since it changes colour between pH 6.8 and pH 5.2 .
(b) The endpoint in this titration would be approximately 6.8 if chlorophenol red were used as an indicator.
(c) The equivalence point in this titration is approximately 7.
(d) The indicator would change from red to yellow.