Chem 105 Chpt 4 Lsn 9 1 CHAPTER 4 Chemical Equations and Stoichiometry Chapter 4 “Quiz” points - complete all assigned homework problems from the chapter - submit on Monday, Feb 12 - problems selected randomly for grading Chapter 4 “Quiz” points - complete all assigned homework problems from the chapter - submit on Monday, Feb 12 - problems selected randomly for grading

Chem 105 Chpt 4 Lsn 9 2 Road Map Where we were Where we were Balanced equations from neutral formulas, containing ionic equations, and one with polyatomic ions (Table 3.1 available on test) Balanced equations from neutral formulas, containing ionic equations, and one with polyatomic ions (Table 3.1 available on test) Relationships: when comparing different compounds, must always go through a MOLE comparison Relationships: when comparing different compounds, must always go through a MOLE comparison Limiting reactant (reagent) Limiting reactant (reagent) Where we are going Where we are going Percent yield Percent yield Chemical equations and chemical analysis Chemical equations and chemical analysis

Chem 105 Chpt 4 Lsn 9 3 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 mL of a M solution of silver nitrate is added to ml of a M solution of sodium chromate, what mass of solid silver chromate (M = g/mol) will be formed? 2 AgNO 3 (aq) + Na 2 CrO 4 (aq) → Ag 2 CrO 4 (s) + 2 NaNO 3 (aq)

Chem 105 Chpt 4 Lsn 9 4 Practice Problem 4-1 Answer 2.00 g Ag 2 CrO 4

Chem 105 Chpt 4 Lsn 9 5 Practice Problem 4-2 A) If mL of a M solution of lead(II) nitrate is added to mL of a M solution of sodium sulfide, what mass of solid lead(II) sulfide will be formed? Pb(NO 3 ) 2 (aq) + Na 2 S(aq) → PbS(s) +2 NaNO 3 (aq)

Chem 105 Chpt 4 Lsn 9 6 Practice Problem 4-2 Answer A) 2.89 g PbS (theoretical) B) Actually make 2.64g (actual) What is your percent yield?

Chem 105 Chpt 4 Lsn 9 7 Practice Problem 4-2B Answer B) Actually make 2.64g What is your percent yield? % yield = × 100 =.9135 x 100 = %

Chem 105 Chpt 4 Lsn 9 8 Practice Problem 4-3 Potassium permanganate reacts with oxalic acid in aqueous sulfuric acid according to the following equation: 2 KMnO H 2 C 2 O H 2 SO 4 → 2 MnSO CO H 2 O + K 2 SO 4 If you start with g of H 2 C 2 O 4 and 75.0 mL of M of KMnO 4 and the percent yield is 85.3 %, what is the actual yield of CO 2 ?

Chem 105 Chpt 4 Lsn 9 9 Practice Problem 4-3 Answers 2.69 g CO 2

Chem 105 Chpt 4 Lsn 9 10 Practice Problem 4-4 Given the chemical reaction between iron and water to form the iron oxide, Fe 3 O 4 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 H 2 O(l) → Fe 3 O 4 (s) + 4 H 2 (g)

Chem 105 Chpt 4 Lsn 9 11 Practice Problem 4-4 Answer 95.6 %

Chem 105 Chpt 4 Lsn 9 12 Practice Problem 4-5 Ammonia is produced by the Haber process using nitrogen and hydrogen gas. If g of nitrogen are reacted with g hydrogen and the reaction yielded g of ammonia. What was the percent yield of the reaction? N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

Chem 105 Chpt 4 Lsn 9 13 Practice Problem 4-5 Answer %

Chem 105 Chpt 4 Lsn 9 14 Chemical Analysis 72. What mass of lime, CaO, can be obtained by heating 125 kg of limestone that is 95% by mass CaCO 3 ? 72. What mass of lime, CaO, can be obtained by heating 125 kg of limestone that is 95% by mass CaCO 3 ? CaCO 3 (s)  CaO (s) + CO 2 (g) 6.65 X 10 4 g CaO 6.65 X 10 4 g CaO

Chem 105 Chpt 4 Lsn 9 15 Chemical Equation A dry-cleaning solvent (M = g/mol) that contains C, H, and Cl is suspected to be a cancer-causing agent. When a g sample was studied by combustion analysis, g of CO 2 and g of H 2 O formed. Calculate the molecular formula. A dry-cleaning solvent (M = g/mol) that contains C, H, and Cl is suspected to be a cancer-causing agent. When a g sample was studied by combustion analysis, g of CO 2 and g of H 2 O formed. Calculate the molecular formula. Empirical C 3 H 2 Cl Empirical C 3 H 2 Cl Molecular C 6 H 4 Cl 2 Molecular C 6 H 4 Cl 2

Chem 105 Chpt 4 Lsn 9 16 Are you up to the challenge? Iodine is made by the reaction Iodine is made by the reaction 2 NaIO 3 (aq) + 5 NaHSO 3 (aq)  3 NaHSO 4 (aq) + 2 Na 2 SO 4 (aq) + H 2 O (l) + I 2 3 NaHSO 4 (aq) + 2 Na 2 SO 4 (aq) + H 2 O (l) + I 2 a) Name the two reactants b) If you wish to prepare 1.00 kg of I 2, what mass of NaIO 3 is required? c) What mass of NaHSO 3 ?

Chem 105 Chpt 4 Lsn 9 17 p.162b

Chem 105 Chpt 4 Lsn 9 18 Next Lesson Chapter 5 Chapter 5 Homework due Homework due

Chem 105 Chpt 4 Lsn 9 19 Balancing Equations ____C 3 H 8 (g) + _____ O 2 (g) ----> _____CO 2 (g) + _____ H 2 O(g) ____B 4 H 10 (g) + _____ O 2 (g) ----> ___ B 2 O 3 (g) + _____ H 2 O(g)

Chem 105 Chpt 4 Lsn 9 20