Chem 105 Chpt 4 Lsn 9Chem 105 Chpt 4 Lsn 9

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CHAPTER 4

Chemical Equations and Stoichiometry

Chapter 4 “Quiz” points - complete all assigned homework problemsfrom the chapter - submit on Monday, Feb 12 - problems selected randomly for grading

Chapter 4 “Quiz” points - complete all assigned homework problemsfrom the chapter - submit on Monday, Feb 12 - problems selected randomly for grading

Chem 105 Chpt 4 Lsn 9Chem 105 Chpt 4 Lsn 9

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Road MapRoad MapWhere we wereWhere we were

Balanced equations from neutral formulas, Balanced equations from neutral formulas, containing ionic equations, and one with containing ionic equations, and one with polyatomic ions (Table 3.1 available on polyatomic ions (Table 3.1 available on test)test)

Relationships: when comparing different Relationships: when comparing different compounds, must always go through a compounds, must always go through a MOLE comparisonMOLE comparison

Limiting reactant (reagent)Limiting reactant (reagent)Where we are goingWhere we are going

Percent yieldPercent yieldChemical equations and chemical analysisChemical equations and chemical analysis

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Practice Problem 4-1

When aqueous silver nitrate and sodium chromate solutions are mixed, a reaction occurs that forms solid silver chromate and a solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of silver nitrate is added to 156.00 ml of a 0.095 M solution of sodium chromate, what mass of solid silver chromate (M = 331.8 g/mol) will be formed?2 AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2 NaNO3(aq)

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Practice Problem 4-1 Answer

2.00 g Ag2CrO4

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Practice Problem 4-2

A) If 257.8 mL of a 0.0468 M solution of lead(II) nitrate is added to 156.00 mL of a 0.095 M solution of sodium sulfide, what mass of solid lead(II) sulfide will be formed?Pb(NO3)2(aq) + Na2S(aq) →

PbS(s) +2 NaNO3(aq)

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Practice Problem 4-2 Answer

A) 2.89 g PbS(theoretical)

B) Actually make 2.64g (actual)

What is your percent yield?

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Practice Problem 4-2B AnswerB) Actually make 2.64gWhat is your percent yield?

% yield = × 100

% yield = × 100

= .9135 x 100 = 91.349 %

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Practice Problem 4-3

Potassium permanganate reacts with oxalic acid in aqueous sulfuric acid according to the following equation:2 KMnO4 + 5 H2C2O4 + 3 H2SO4 →

2 MnSO4 + 10 CO2 + 8 H2O + K2SO4

If you start with 3.225 g of H2C2O4 and 75.0 mL of 0.250 M of KMnO4 and the percent yield is 85.3 %, what is the actual yield of CO2?

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Practice Problem 4-3 Answers

2.69 g CO2

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Practice Problem 4-4

Given the chemical reaction between iron and water to form the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is reacted with sufficient water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed?3 Fe(s) + 4 H2O(l) → Fe3O4(s) + 4 H2(g)

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Practice Problem 4-4 Answer

95.6 %

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Practice Problem 4-5

Ammonia is produced by the Haber process using nitrogen and hydrogen gas. If 85.90 g of nitrogen are reacted with 21.66 g hydrogen and the reaction yielded 98.67 g of ammonia. What was the percent yield of the reaction?

N2(g) + 3 H2(g) → 2 NH3(g)

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Practice Problem 4-5 Answer

94.49 %

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Chemical AnalysisChemical Analysis

72. What mass of lime, CaO, can be 72. What mass of lime, CaO, can be obtained by heating 125 kg of obtained by heating 125 kg of limestone that is 95% by mass limestone that is 95% by mass CaCOCaCO33??

CaCOCaCO33 (s) (s) CaO (s) + CO CaO (s) + CO22 (g) (g)

6.65 X 106.65 X 1044 g CaO g CaO

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Chemical EquationChemical Equation

A dry-cleaning solvent (A dry-cleaning solvent (MM = 146.99 = 146.99 g/mol) that contains C, H, and Cl is g/mol) that contains C, H, and Cl is suspected to be a cancer-causing suspected to be a cancer-causing agent. When a 0.250 g sample was agent. When a 0.250 g sample was studied by combustion analysis, 0.451 studied by combustion analysis, 0.451 g of COg of CO22 and 0.0617 g of H and 0.0617 g of H22O formed. O formed. Calculate the molecular formula.Calculate the molecular formula.

Empirical CEmpirical C33HH22ClCl

Molecular CMolecular C66HH44ClCl22

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Are you up to the challenge?Are you up to the challenge? Iodine is made by the reactionIodine is made by the reaction

2 NaIO2 NaIO33 (aq) + 5 NaHSO (aq) + 5 NaHSO33 (aq) (aq)

3 NaHSO3 NaHSO44 (aq) + 2 Na (aq) + 2 Na22SOSO44 (aq) + H (aq) + H22O (l) O (l) + I+ I22

a)a) Name the two reactantsName the two reactants

b)b) If you wish to prepare 1.00 kg of IIf you wish to prepare 1.00 kg of I22, , what mass of NaIOwhat mass of NaIO33 is required? is required?

c)c) What mass of NaHSOWhat mass of NaHSO33??

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p.162b

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Next LessonNext Lesson

Chapter 5Chapter 5Homework dueHomework due

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Balancing Balancing EquationsEquationsBalancing Balancing EquationsEquations

____C____C33HH88(g) + _____ O(g) + _____ O22(g) ---->(g) ---->

_____CO_____CO22(g) + _____ H(g) + _____ H22O(g)O(g)

____B____B44HH1010(g) + _____ O(g) + _____ O22(g) ---->(g) ---->

___ B___ B22OO33(g) + _____ H(g) + _____ H22O(g)O(g)

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