Types of Chemical Reactions and Solution Stoichiometry Chapter 4 E-mail:...

Types of Chemical Reactions and Solution Stoichiometry

Chapter 4E-mail: [email protected]

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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?


Dissolved ionic (NaCl) Dissolved molecules (Sugar)

Electrolyte Solution

Non-electrolyte Solution

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. CH2O

c. HNO3

d. H2S


Lots of ions – Conducts well

Few ions – Conducts weakly

No ions – Does not conduct


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 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?


Molarity (M) a unit of concentration ⇒that measures the moles of solute

per liter of solutionM =

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?

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?


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 48. 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 49. 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 410. 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


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:

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


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

Types of Chemical Reactions and Solution Stoichiometry – ch 412. Balance the following red-ox reactions:

a. MnO2 + NO3- N2O4 + MnO4

- (acidic)

b. CNO- + As2O3 CN- + HAsO42- (basic)

c. Cr3+ + SO42– Cr2O7

2– + H2SO3 (acidic)


Balancing Red-ox Reactions1. Separate into two half reactions2. Balance all elements except H and O3. Balance the oxygen by adding H2O

4. 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 H2O

8. 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.

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 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)


You have completed ch. 4

Answer Key – Ch. 41. 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. 43. 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⇒ ⇒

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+

0.38 mol/L BaF2 2 mol F– = 0.76 M F–

Answer Key – Ch. 4

1 mol BaF2

1 mol BaF2

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–


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. 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 reaction

2 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)/3V1 = 356 mL of Li3PO4

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 ions

Ag+ (aq) + Cl– (aq) AgCl (s) b. How many grams of precipitate form?

First determine the LR

…continue to next slide

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 LR

3.2 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


1 mol AgNO3 1 mol AgCl

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+


V1 = 160 mL


Coeff Coeff

(0.15 M)(V1)/1 = (0.8 M)(60 mL)/2

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

Answer Key – Ch. 410. 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. 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⇒ ⇒

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 agent

Li is getting oxidized ⇒ reducing agent

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

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. 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 HAsO42– + 3 H2O + 4 e–

2 CNO– + As2O3 + 4 OH– 2 CN– + 2 HAsO42– + H2O


Answer Key – Ch. 412. …continued

c. Cr3+ + SO42– Cr2O7

2– + 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 + CrO7

2–

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 = 0.210 mole

HCl ⇒ (325mL)(1.15g/mL)(0.301) = 112.5g = 3.08 mole



Answer Key – Ch. 413. ….continued

= vs. = 0.513

Limiting reagent⇒ K2Cr2O7

0.210 K2Cr2O7 = 44.67 g

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