Types of Chemical Reactions and Solution Stoichiometry
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Transcript of Types of Chemical Reactions and Solution Stoichiometry
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Types of Chemical Reactions and Solution StoichiometryChapter 4E-mail: [email protected]: http://clas.sa.ucsb.edu/staff/terri/
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Types of Chemical Reactions and Solution Stoichiometry – ch 41. Distinguish between solute, solvent and solution.
2. What types of solutes are electrolytes?
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Types of Chemical Reactions and Solution Stoichiometry – ch 4
Dissolved ionic (NaCl) Dissolved molecules (Sugar)
Electrolyte Solution
Non-electrolyte Solution
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Types of Chemical Reactions and Solution Stoichiometry – ch 43. Determine if the following substances are strong, weak or
non-electrolytes in water.a. KC2H3O2
b. CH2Oc. HNO3
d. H2S
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Types of Chemical Reactions and Solution Stoichiometry – ch 4
Lots of ions – Conducts well
Few ions – Conducts weakly
No ions – Does not conduct
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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
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Types of Chemical Reactions and Solution Stoichiometry – ch 44. 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?
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Types of Chemical Reactions and Solution Stoichiometry – ch 45. 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?
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Types of Chemical Reactions and Solution Stoichiometry – ch 46. 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?
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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 reactionb. How many grams of precipitate form?c. What is the concentration of the ions in the solution after the
reaction has gone to completion?
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Types of Chemical Reactions and Solution Stoichiometry – ch 48. What type of solution results (acidic, basic or neutral) when 110 mL
of 0.40 M HC2H3O2 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the molecular and net ionic equations.
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Types of Chemical Reactions and Solution Stoichiometry – ch 49. What volume of 0.15 M H3PO4 solution is required to neutralize 60
mL of 0.80 M KOH solution?
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Types of Chemical Reactions and Solution Stoichiometry – ch 410. Assign oxidation states for the following:
a. S in SO3
b. P in PF5
c. Cl in Cl2
d. C in C3H8
e. Mn in MnO4–
f. Cr in K2Cr2O7
g. N in NH4NO3
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Types of Chemical Reactions and Solution Stoichiometry – ch 4
Hierarchy of Substances Oxidation State
Hydrogen in a molecular compound or polyatomic ion
+1
Fluorine in a molecular compound -1Oxygen in a molecular compound or
polyatomic ion-2
Chlorine, bromine or iodine in a molecular compound
-1
Nitrogen in a molecular compound or polyatomic ion
-3
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 ion2. The hierarchy tells which element gets assigned first:
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Types of Chemical Reactions and Solution Stoichiometry – ch 411. 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
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Types of Chemical Reactions and Solution Stoichiometry – ch 4
Oxidation-Reduction (Redox) Reactions1. A reaction in which there is a transfer of electrons2. Oxidation is when an element loses electrons and is noted by an
increase in the oxidation number3. Reduction is when an element gains electrons and is noted by a
decrease in the oxidation number4. The number of electrons gained must equal the number of electrons
lost5. An oxidizing agent is the substance getting reduced6. A reducing agent is the substance getting oxidized
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Types of Chemical Reactions and Solution Stoichiometry – ch 412. Balance the following red-ox reactions:
a. Cr3+ + SO42– Cr2O7
2– + H2SO3 (acidic)b. CNO- + As2O3 CN- + HAsO4
2- (basic)
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Types of Chemical Reactions and Solution Stoichiometry – ch 4
Balancing Red-ox Reactions1. Separate into two half reactions2. Balance all elements except H and O3. Balance the oxygen by adding H2O4. Balance the hydrogen by adding H+
5. Balance the net charges by adding electrons (e-) to the more positive side6. Make the electrons lost equal to the electrons gained by multiplying the half
reactions by the smallest common multiple7. Add two half reactions back together canceling out electrons, H+ and H2O8. If acidic stop. If basic add an OH- to each side for every H+. The OH-
cancels out the H+ making water, which then needs to be adjusted for.
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Types of Chemical Reactions and Solution Stoichiometry – ch 413. 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 89.5% K2Cr2O7 by mass is allowed to react with 225 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)
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Types of Chemical Reactions and Solution Stoichiometry – ch 4
You have completed ch. 4
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Answer Key – Ch. 41. Distinguish between solute, solvent and solution.
solute – a substance dissolved (typically into a liquid) to form a solutionsolvent – the dissolving medium in a solutionsolution – 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
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Answer Key – Ch. 43. Determine if the following substances are strong, weak or non
electrolytes in water.a. KC2H3O2 ⇒ soluble ionic ⇒ strong electrolyteb. CH2O ⇒ molecular ⇒ non electrolytec. HNO3 ⇒ strong acid ⇒ strong electrolyted. H2S weak acid weak electrolyte⇒ ⇒
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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 LM = (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+
0.38 mol/L BaF2 2 mol F– = 0.76 M F–
Answer Key – Ch. 4
1 mol BaF2
1 mol BaF2
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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
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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
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Answer Key – Ch. 46. 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 reaction2 Li3PO4 (aq) + 3 Hg(NO3)2 (aq) 6 LiNO3 (aq) + Hg3(PO4)2 (s)
M1V1 = M2V2coeff coeff(0.15 M)(V1)/2 = (0.32 M)(250 mL)/3
V1 = 356 mL of Li3PO4
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Answer Key – Ch. 47. 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 ionsAg+ (aq) + Cl– (aq) AgCl (s)
b. How many grams of precipitate form?First determine the LR
…continue to next slide
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7. …continuedCaCl2 ⇒ (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 LR3.2 mmols AgNO3 1mol AgCl 143.32 g AgCl = 459 mg 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–
or 0.459 g AgCl
Answer Key – Ch. 4
1 mol AgNO3 1 mol AgCl
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8. What type of solution results (acidic, basic or neutral) when 110 mL of 0.40 M HC2H3O2 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the molecular and net ionic equations.
Molecular ⇒ 2 HC2H3O2 (aq) + Ba(OH)2 (aq) 2 H2O (l) + Ba(NO3)2 (aq)Net Ionic ⇒ HC2H3O2 (aq) + OH– (aq) H2O (l) + C2H3O2
– (aq)
H+ ⇒ (110 mL)(0.4 M) = 44 mmolsOH– ⇒ (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
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V1 = 107 mL
Answer Key – Ch. 4
Coeff Coeff
(0.15 M)(V1)/1 = (0.8 M)(60 mL)/3
9. What volume of 0.15 M H3PO4 is required to neutralize 60 mL of 0.80 M KOH?
H3PO4 + 3 KOH 3 H2O + K3PO4
M1V1 = M2V2
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Answer Key – Ch. 410. Assign oxidation states for the following:
a. S in SO3 ⇒ S + 3(-2) = 0 ⇒ S = +6b. P in PF5 ⇒ P + 5(-1) = 0 ⇒ P = +5c. Cl in Cl2 Cl = 0⇒d. Mn in MnO4
– (+1) + Mn + 4(-2) = 0 Mn = +7⇒ ⇒
e. Cr in K2Cr2O7 ⇒ 2(Cr) + 7(-2) = -2 Cr = +6⇒
f. N in NH4NO3 ⇒ In NH4+ N + 4(+1) = +1 N = -3⇒ ⇒
In NO3– N + 3(-2) = -1 N = +5⇒ ⇒
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Answer Key – Ch. 411. 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 agentLi is getting oxidized ⇒ reducing agent
b. 3 Cu + 2 NO3– + 8 H+ 3 Cu2+ + 2 NO + 4 H2O
Cu is getting oxidized reducing agentN in NO3
– is getting reduced oxidizing agent
c. Pb + PbO2 + 2 H2SO4 2 PbSO4 + 2 H2OPb is getting oxidized reducing agent
Pb in PbO2 is getting reduced oxidizing agent
2 moles of electrons
6 moles of electrons
2 moles of electrons
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Answer Key – Ch. 412. 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 HAsO4
2– + 3 H2O + 4 e– 2 CNO– + As2O3 + 4 OH– 2 CN– + 2 HAsO4
2– + H2O
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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 89.5% K2Cr2O7 by mass is allowed to react with 225 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+ + Cr2O7
2– + 6 Cl- 7 H2O + 2 Cr3+ + 3 Cl2
⇒ (0.895)(62.6g)()()= 0.19 mole H+ ⇒ (225mL)(1.15g/mL)(0.301)()()= 2.13 mole H+
…continue to next slide
Answer Key – Ch. 4
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Answer Key – Ch. 413. ….continued = vs. = 0.153Limiting reagent⇒ H+
0.153 mol H+()( )( ) = 44.67 g