Types of Chemical Reactions and Solution Stoichiometry

Report
Types of Chemical Reactions
and Solution Stoichiometry
Chapter 4
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Types of Chemical Reactions and Solution
Stoichiometry – ch 4
1. Distinguish between solute, solvent and solution.
2. What types of solutes are electrolytes?
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
Dissolved ionic (NaCl)
Electrolyte
Solution
Dissolved molecules (Sugar)
Non-electrolyte
Solution
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
3. Determine if the following substances are strong, weak or
non-electrolytes in water.
a. KC2H3O2
b. CH2O
c. HNO3
d. H2S
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
Lots of ions –
Conducts well
Few ions –
Conducts weakly
No ions –
Does not conduct
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
Type of compound
Strong
Weak
Non
Ionic
Only soluble
Slightly or insoluble
N/A
Acids
HCl, HBr, HI,
HNO3, HClO3,
HClO4, H2SO4
HBrO4, HIO4
All other acids
N/A
Molecular
N/A
NH3
All other
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
4. A solution is prepared by dissolving 50 g of barium fluoride in
enough water to make 750 mL of solution.
a. What is the molarity of the barium fluoride?
b. What is the concentration for each of the ions in solution?
c. What is the concentration of the fluoride ion if 350 mL of water
is added?
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
Molarity (M) ⇒ a unit of concentration
that measures the moles of solute
per liter of solution
nsolute
M=
Lsolution
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
5. What is the resulting concentration of bromide ion if 28 mL of 0.15
M sodium bromide solution were mixed with 45 mL of 0.12 M
calcium bromide solution?
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
6. What volume of 0.150 M Li3PO4 in milliliters is required to
precipitate all of the mercury(II) ions from 250 mL of 0.32 M
Hg(NO3)2?
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
7. A precipitate will form when 10 mL of 0.25 M calcium chloride is mixed
with 8 mL of 0.4 M silver nitrate.
a. Write the molecular and net ionic equation for this reaction
b. How many grams of precipitate form?
c. What is the concentration of the ions remaining in solution?
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
8. What type of solution results (acidic, basic or neutral) when 110 mL
of 0.40 M HNO3 is mixed with 42 mL of 0.90 M Ba(OH)2? Write
the molecular and net ionic equations.
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
9. What volume of 0.15 M H2SO4 is required to neutralize 60 mL of
0.80 M KOH?
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
10. Assign oxidation states for the following:
a. S in SO3
b. P in PF5
c. Cl in Cl2
d. C in C3H8
d. Mn in MnO4–
e. Cr in K2Cr2O7
f. N in NH4NO3
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
Rules and Hierarchy for Assigning Oxidation Numbers/States
1. What type of substance are you analyzing?
a. Element ⇒ The oxidation state for atoms in their elemental form are 0
b. Molecular Compound ⇒ the sum of the oxidation numbers in a compound
equals 0
c. Ionic compound ⇒ analyze the cation and anion individually
- Mono-atomic ions ⇒ The oxidation state for mono-atomic ions is the
charge on the ion
- Polyatomic ions ⇒ The sum of the oxidation numbers in a polyatomic ion
equals the net charge on the ion
2. The hierarchy tells which element gets assigned first:
Hierarchy of Substances
Oxidation State
Hydrogen in a molecular compound or polyatomic ion
Fluorine in a molecular compound
Oxygen in a molecular compound or polyatomic ion
Chlorine, bromine or iodine in a molecular compound
Nitrogen in a molecular compound or polyatomic ion
+1
-1
-2
-1
-3
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
11. Identify the oxidizing and reducing agents for the following
reactions and determine the moles of electrons transferred:
a. Ba2+ + 2 Li  Ba + 2 Li+
b. 3 Cu + 2 NO3– + 8 H+  3 Cu2+ + 2 NO + 4 H2O
c. Pb + PbO2 + 2 H2SO4  2 PbSO4 + 2 H2O
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
1.
2.
3.
4.
5.
6.
Oxidation-Reduction (Redox) Reactions
A reaction in which there is a transfer of electrons
Oxidation is when an element loses electrons and is noted by an
increase in the oxidation number
Reduction is when an element gains electrons and is noted by a
decrease in the oxidation number
The number of electrons gained must equal the number of electrons
lost
An oxidizing agent is the substance getting reduced
A reducing agent is the substance getting oxidized
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
12. Balance the following red-ox reactions:
a. MnO2 + NO3-  N2O4 + MnO4(acidic)
b. CNO- + As2O3  CN- + HAsO42- (basic)
c. Cr3+ + SO42–  Cr2O72– + H2SO3
(acidic)
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
1.
2.
3.
4.
5.
6.
7.
8.
Balancing Red-ox Reactions
Separate into two half reactions
Balance all elements except H and O
Balance the oxygen by adding H2O
Balance the hydrogen by adding H+
Balance the net charges by adding electrons (e-) to the more positive side
Make the electrons lost equal to the electrons gained by multiplying the half
reactions by the smallest common multiple
Add two half reactions back together canceling out electrons, H+ and H2O
If acidic stop. If basic add an OH- to each side for every H+. The OHcancels out the H+ making water, which then needs to be adjusted for.
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
13. The reaction below can be used as a laboratory method of preparing
small quantities of Cl2(g). If a 62.6 g sample that is 98.5% K2Cr2O7
by mass is allowed to react with 325 mL of HCl(aq) with a density
of 1.15 g/mL and 30.1% HCl by mass, how many grams of Cl2(g)
are produced?
Cr2O72- (aq) + Cl-(aq)  Cr3+(aq) + Cl2(g) (unbalanced)
Types of Chemical Reactions and Solution
Stoichiometry – ch 4
You have completed ch. 4
Answer Key – Ch. 4
1. Distinguish between solute, solvent and solution.
solute – a substance dissolved (typically into a liquid) to form a
solution
solvent – the dissolving medium in a solution
solution – homogenous mixture of solute and solvent or 2 solvents
2. What is an electrolyte?
electrolyte – a substance that produces ions upon dissolving in
water thus causing a solution that can conduct electricity – all
ionic compounds and acids are electrolytes – only some molecular
are electrolytes
Answer Key – Ch. 4
3. Determine if the following substances are strong, weak or non
electrolytes in water.
a. KC2H3O2 ⇒ soluble ionic ⇒ strong electrolyte
b. CH2O ⇒ molecular ⇒ non electrolyte
c. HNO3 ⇒ strong acid ⇒ strong electrolyte
d. H2S ⇒ weak acid ⇒ weak electrolyte
Answer Key – Ch. 4
4. A solution is prepared by dissolving 50 g of barium fluoride in
enough water to make 750 mL of solution.
a. What is the molarity of the barium fluoride?
(50 g BaF2)/(175.33 g/mol) = 0.285 mol BaF2
(750 mL)(1L/1000mL) = 0.75 L
M = (0.285 mol)/(0.75 L) = 0.38 M BaF2
b. What is the concentration for each of the ions in solution?
BaF2 (s)  Ba2+ (aq) + 2F– (aq)
0.38 mol/L BaF2 1 mol Ba2+ = 0.38 M Ba2+
1 mol BaF2
0.38 mol/L BaF2 2 mol F–
= 0.76 M F–
1 mol BaF2
Answer Key – Ch. 4
c. What is the concentration of the fluoride ion if 350 mL of water
is added?
since the volume will go up the solution will get diluted – if
you’re only adding solvent the moles of solute will remain
constant ⇒ M1V1 = M2V2
(0.76 M F–)(750 mL) = (M2)(750 mL + 350 mL)
M2 = 0.52 M F–
Answer Key – Ch. 4
5. What is the resulting concentration of bromide ion if 28 mL of 0.15 M
sodium bromide solution were mixed with 45 mL of 0.12 M calcium
bromide solution?
NaBr (s)  Na+ (aq) + Br– (aq)
0.15 M NaBr ⇒ 0.15 M Br–
(0.15 mol/L)(28 mL) = 4.2 mmol Br– (before mixing)
CaBr2 (s)  Ca2+ (aq) + 2 Br– (aq)
0.12 M CaBr2 ⇒ 0.24 M Br–
(0.24 mol/L)(45 mL) = 10.8 mmol Br– (before mixing)
After mixing ⇒ 4.2 mmol Br– + 10.8 mmol = 15 mmol Br–
Final concentration ⇒ (15 mmol Br– )/(28 mL + 45 mL)
= 0.21 M Br–
Answer Key – Ch. 4
6. What volume of 0.150 M Li3PO4 in milliliters is required to
precipitate all of the mercury(II) ions from 250 mL of 0.32 M
Hg(NO3)2?
First write a balanced chemical reaction
2 Li3PO4 (aq) + 3 Hg(NO3)2 (aq)  6 LiNO3 (aq) + Hg3(PO4)2 (s)
M1V1 = M2V2
coeff
coeff
(0.15 M)(V1)/2 = (0.32 M)(250 mL)/3
V1 = 356 mL of Li3PO4
Answer Key – Ch. 4
7. A precipitate will form when 10 mL of 0.25 M calcium chloride is mixed
with 8 mL of 0.4 M silver nitrate.
a. Write the molecular and net ionic equation for this reaction
Molecular ⇒
CaCl2 (aq) + 2 AgNO3 (aq)  Ca(NO3)2 (aq) + 2 AgCl (s)
Net ionic ⇒
Ca2+ and NO3– are the spectator ions
Ag+ (aq) + Cl– (aq)  AgCl (s)
b. How many grams of precipitate form?
First determine the LR
…continue to next slide
Answer Key – Ch. 4
7. …continued
CaCl2 ⇒ (10 mL)(0.25 M) = 2.5 mmols/1 = 2.5
AgNO3 ⇒ (8 mL)(0.4 M) = 3.2 mmols/2 = 1.6 AgNO3 is the LR
3.2 mmols AgNO3 1mol AgCl 143.32 g AgCl = 459 mg AgCl
1 mol AgNO3 1 mol AgCl
or 0.459 g AgCl
c. What is the concentration of the ions remaining in solution?
Spectator ions don’t get consumed ⇒
[Ca2+] = (2.5 mmol)/(10 mL + 8 mL) = 0.14 M Ca2+
[NO3–] = (3.2 mmol)/(18 mL) = 0.18 M NO3–
Reacting ions get consumed ⇒
[Ag+] = (3.2 mmol – 3.2 mmol) = 0 M (consumed entirely b/c it’s the LR)
[Cl–] = (5 mmol – 3.2 mmol)/(18 mL) = 0.1 M Cl–
Answer Key – Ch. 4
8. What type of solution results (acidic, basic or neutral) when 110 mL of
0.40 M HNO3 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the
molecular and net ionic equations.
Molecular ⇒ 2 HNO3 (aq) + Ba(OH)2 (aq)  2 H2O (l) + Ba(NO3)2 (aq)
Net Ionic ⇒ H+ (aq) + OH– (aq) H2O (l)
H+ ⇒ (110 mL)(0.4 M) = 44 mmols
OH– ⇒ (42 mL)(1.8 M) = 75.6 mmols
The solution will be basic because the moles of OH – exceed the moles of H+
Answer Key – Ch. 4
9. What volume of 0.15 M H2SO4 is required to neutralize 60 mL of
0.80 M KOH?
H2SO4 + 2 KOH  2 H2O + K2SO4
M1V1 = M2V2
Coeff
Coeff
(0.15 M)(V1)/1 = (0.8 M)(60 mL)/2
V1 = 160 mL
Answer Key – Ch. 4
10. Assign oxidation states for the following:
a. S in SO3 ⇒ S + 3(-2) = 0 ⇒ S = +6
b. P in PF5 ⇒ P + 5(-1) = 0 ⇒ P = +5
c.
d.
e.
f.
Cl in Cl2 ⇒ Cl = 0
Mn in MnO4– ⇒ (+1) + Mn + 4(-2) = 0 ⇒ Mn = +7
Cr in K2Cr2O7 ⇒ 2(Cr) + 7(-2) = -2 ⇒ Cr = +6
N in NH4NO3 ⇒ In NH4+ ⇒ N + 4(+1) = +1 ⇒ N = -3
In NO3– ⇒ N + 3(-2) = -1 ⇒ N = +5
Answer Key – Ch. 4
11. Identify the oxidizing and reducing agents for the following
reactions and determine the moles of electrons transferred:
a. Ba2+ + 2 Li  Ba + 2 Li+
Ba2+ is getting reduced ⇒ oxidizing agent
Li is getting oxidized ⇒ reducing agent
2 moles of electrons
b. 3 Cu + 2 NO3– + 8 H+  3 Cu2+ + 2 NO + 4 H2O
Cu is getting oxidized reducing agent
N in NO3– is getting reduced oxidizing agent
6 moles of electrons
c. Pb + PbO2 + 2 H2SO4  2 PbSO4 + 2 H2O
Pb is getting oxidized reducing agent
Pb in PbO2 is getting reduced oxidizing agent
2 moles of electrons
Answer Key – Ch. 4
12. Balance the following red-ox reactions:
a. MnO2 + NO3-  N2O4 + MnO4- (acidic)
Reduction ½ : 2x(3 e– + 2 H2O + MnO2  MnO4– + 4 H+)
Oxidation ½ : 3x(4 H+ + 2 NO3–  N2O4 + 2H2O + 2 e–)
2 MnO2 + 6 NO3– 4 H+  2 MnO4– + 3 N2O4 + 2 H2O
b. CNO- + As2O3  CN- + HAsO42- (basic)
Reduction ½ : 2x(2 e– + H2O + CNO–  CN– + 2 OH–)
Oxidation ½ : 8 OH– + As2O3  2 HAsO42– + 3 H2O + 4 e–
2 CNO– + As2O3 + 4 OH–  2 CN– + 2 HAsO42– + H2O
…continue to next slide
Answer Key – Ch. 4
12. …continued
c. Cr3+ + SO42–  Cr2O72– + H2SO3
(acidic)
Reduction ½ : 3x(2 e– + 4 H+SO42-  H2SO3 + H2O)
Oxidation ½ : 7 H2O + 2 Cr3+  6 e– + 14 H+ + Cr2O72–
3 SO42– + 2 Cr3+ + 4 H2O  2 H+ + 3 H2SO3 + CrO72–
Answer Key – Ch. 4
13. The reaction below can be used as a laboratory method of preparing
small quantities of Cl2(g). If a 62.6 g sample that is 98.5% K2Cr2O7
by mass is allowed to react with 325 mL of HCl(aq) with a density
of 1.15 g/mL and 30.1% HCl by mass, how many grams of Cl2(g)
are produced?
Cr2O72- (aq) + Cl-(aq)  Cr3+(aq) + Cl2(g) (unbalanced)
Oxidation ½ : 6 e– + 14 H+ + Cr2O72– 7 H2O + 2 Cr3+
Reductions ½ : 3x(2 Cl-  Cl2 + 2 e-)
14 H+ + Cr2O72– + 6 Cl-  7 H2O + 2 Cr3+ + 3 Cl2
K2Cr2O7 ⇒ (0.985)(62.6g) = 61.7g 1mole = 0.210 mole
294.2 g
HCl ⇒ (325mL)(1.15g/mL)(0.301) = 112.5g 1mole = 3.08 mole
36.461 g
…continue to next slide
Answer Key – Ch. 4
13. ….continued
0.210 mole K2Cr2O7 = 0.210 vs. 3.08 mole = 0.513
1
6
Limiting reagent⇒ K2Cr2O7
2–
1mol
Cr
O
3 mol Cl2 70.906 g Cl2= 44.67 g Cl
2
7
0.210 K2Cr2O7
2
1 mol K2Cr2O7 1mol Cr2O72– 1 mol Cl2

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