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 1 Chem 105 Chpt 4 Lsn 9

Road Map

 Where we were    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 Limiting reactant (reagent)  Where we are going  Percent yield  Chemical equations and chemical analysis Chem 105 Chpt 4 Lsn 9 2

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 AgNO 3 (aq) + Na 2 CrO 4 (aq) → Ag 2 CrO 4 (s) + 2 NaNO 3 (aq) 3 Chem 105 Chpt 4 Lsn 9

Practice Problem 4-1 Answer

 2.00 g Ag 2 CrO 4 Chem 105 Chpt 4 Lsn 9 4

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(NO 3 ) 2 (aq) + Na 2 S(aq) → PbS(s) +2 NaNO 3 (aq) 5 Chem 105 Chpt 4 Lsn 9

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 6

Practice Problem 4-2B Answer

 B) Actually make 2.64g

What is your percent yield?

% yield = × 100 % yield = × 100 = .9135 x 100 = 91.349 % Chem 105 Chpt 4 Lsn 9 7

Practice Problem 4-3

 Potassium permanganate reacts with oxalic acid in aqueous sulfuric acid according to the following equation: 2 KMnO 4 + 5 H 2 C 2 O 4 2 MnSO 4 + 3 H + 10 CO 2 2 SO 4 + 8 H → 2 O + K 2 SO 4 If you start with 3.225 g of H mL of 0.250 M of KMnO 4 2 C 2 O 4 and 75.0 CO 2 ?

and the percent yield is 85.3 %, what is the actual yield of 8 Chem 105 Chpt 4 Lsn 9

Practice Problem 4-3 Answers

 2.69 g CO 2 Chem 105 Chpt 4 Lsn 9 9

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) 10 Chem 105 Chpt 4 Lsn 9

Practice Problem 4-4 Answer

 95.6 % Chem 105 Chpt 4 Lsn 9 11

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?

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) 12 Chem 105 Chpt 4 Lsn 9

Practice Problem 4-5 Answer

 94.49 % Chem 105 Chpt 4 Lsn 9 13

Chemical Analysis

 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 14 Chem 105 Chpt 4 Lsn 9

Chemical Equation

 A dry-cleaning solvent ( analysis, 0.451 g of CO 2 M = 146.99 g/mol) that contains C, H, and Cl is suspected to be a cancer-causing agent. When a 0.250 g sample was studied by combustion and 0.0617 g of H 2 O formed. Calculate the molecular formula.

  Empirical C 3 H 2 Cl Molecular C 6 H 4 Cl 2 15 Chem 105 Chpt 4 Lsn 9

Are you up to the challenge?

 Iodine is made by the reaction 2 NaIO 3 (aq) + 5 NaHSO 3 (aq) a) 3 NaHSO 4 (aq) + 2 Na 2 SO 4 Name the two reactants  (aq) + H 2 O (l) + I 2 b) c) If you wish to prepare 1.00 kg of I 2 , what mass of NaIO 3 is required?

What mass of NaHSO 3 ?

16 Chem 105 Chpt 4 Lsn 9

Chem 105 Chpt 4 Lsn 9 17

p.162b

Next Lesson

 Chapter 5  Homework due Chem 105 Chpt 4 Lsn 9 18

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 19

Chem 105 Chpt 4 Lsn 9 20