Chapter 9 Objectives:To understand the molecular and mass information given in a balanced equation.

NOW LET’S GET BACK to… CO(g) + 2H 2 (g) CH 3 OH(l)

9.2 Mole-Mole Relationships Objectives: To learn to use a balanced equation to determine relationships between moles of reactants and moles of products.

Mole-Ratio: 2H 2 O 2H 2 (g) + O 2 (g) 4H 2 O 4H 2 (g) + 2O 2 (g) 5.8H 2 O 5.8H 2 (g) + ?O 2 (g)

Mole-Ratio: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) What number of moles of oxygen gas is required to react with 7.4 mol of methane? How many moles of carbon dioxide gas will be produced by reacting 2.6 mol of oxygen with excess methane?

Mole-Ratio: 2H 2 S(g)+3O 2 (g) 2SO 2 (g) + 2H 2 O How many moles of oxygen gas are required to react with 5.6 mol of hydrogen sulfide? How many moles of sulfur dioxide gas will be produced by reacting 7.3 mol of hydrogen sulfide with excess oxygen? How many moles of sulfur dioxide gas will be produced by reacting 7.3 mol of oxygen with excess hydrogen sulfide?

Mole-Ratio: HW SC9.1 4FeO(s) + O 2 (g) 2Fe 2 O 3 (s) What number of moles of oxygen gas is required to react with 2.4 mol of iron(II) oxide? What number of moles of iron(III) oxide will be produced by reacting 9.2 mol of iron(II)oxide with excess oxygen? What number of moles of iron(III) oxide will be produced by reacting 9.2 mol of oxygen with excess iron(II) oxide?

9.3 Mass Calculations Objective: To learn to relate masses of reactants and products in a chemical reaction.

9.3 Mass Calculations BACK TO THE DELI 2 pieces of bread+3 slices of meat+ 1 cheese 1 sandwich If each slice of bread weighs ½ of a slice of meat and the total meat = 6 oz, what is the weight of bread necessary for 1 sandwich?

9.3 Mass Calculations C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g). What mass of oxygen is needed to react exactly with 44.1 g of propane? Step1: Balance the equation C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g).

9.3 Mass Calculations C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) g? g O 2 ? moles? mol O g C 3 H 8 X 1mol C 3 H 8 = 1.00 mol C 3 H g C 3 H 8 Step 2 5 mol of O 2 = 1molC 3 H 8 Step 3 5 mol x 32.0g O 2 1 mol ANSWER

9.3 Mass Calculations 2NaOH(aq)+CO 2 (g) Na 2 CO 3 (g) + H 2 O(l) 10 g? g CO 2 ? moles? mol CO 2 Step 2 Step 3 ANSWER 5.5 g CO 2

9.3 Mass Calculations 10 g? g CO 2 ? moles? mol CO 2 Step 2 Step 3 ANSWER 5.5 g CO 2 Homework: Self-check exercises 9.2, 9.3, and 9.4

9.5: Mass Calculations-Comparing 2 Reactions Objectives: To compare the stoichiometry of two reactions Which is better……………… Baking soda or Milk of Magnesia

9.5: Mass Calculations-Comparing 2 Reactions How many moles of HCl will react with 1.00 g of each antacid? Baking soda NaHCO 3 (s) + HCl(aq) NaCl(aq)+ H 2 O + CO 2 (g) Milk of Magnesia Mg(OH) 2 (s) + 2HCl(aq) 2H 2 O + MgCl 2 (aq)

Section 9.4: Mass Calculations using Scientific Notation Objectives: To carry out mass calculations that involve scientific notation. Stoichiometry: process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.

Section 9.4: Mass Calculations using Scientific Notation 1)Divide into 3 or 4 teams p.264 2)Using the equation provided calculate the mass of the substance produced if you use up all of that Top Ten chemical. Ca 3 (PO4) 2 + H 2 SO4 H 3 PO 4 N 2 + H 2 NH 3 C 2 H 4 + O 2 CO 2 + H 2 O CaCO 3 CaO+CO 2 NaOH + H 3 PO 4 Na 3 PO 4 + H 2 O C 3 H 7 + O 2 CO 2 + H 2 O Na + Cl 2 NaCl

9.6: The Concept of Limiting Reactants Objective: To understand the concept of limiting reactants

9.6: The Concept of Limiting Reactants Deli: 20 slices of bread 24 slices of meat 12 slices of cheese 2 pieces of bread + 3 meat + 1 cheese How many sandwiches? What is limiting? NCSSM Distance Learning T.I.G.E.R. - Chemistry page 2#stoich

Before the Reaction

Before and after the reaction.

Example #2

Before and after the reaction. Limiting reactant

NCSSM Distance Learning T.I.G.E.R. - Chemistry page 2#stoichNCSSM Distance Learning T.I.G.E.R. - Chemistry page 2#stoich

Section 9.7: Calculations involving a Limiting Reactant Objectives:To learn to recognize the limiting reactant in a reaction. To learn to use the limiting reactant to do stoichiometric calculations.

Section 9.7: Calculations involving a Limiting Reactant Objectives:To learn to recognize the limiting reactant in a reaction. To learn to use the limiting reactant to do stoichiometric calculations.

9.7 Limiting Reactant CH 4 (g) + H 2 O(g) 3H 2 (g) + CO(g) 249 g? g H 2 O ? moles? mol H 2 O Step 2 Step 3 ANSWER 279g H2O 15.5 mol

9.7 Limiting Reactant CH 4 (g) + H 2 O(g) 3H 2 (g) + CO(g) 249 g300 g H 2 O ? moles? mol H 2 O

9.7 Limiting Reactant CH 4 (g) + H 2 O(g) 3H 2 (g) + CO(g) 249 g300 g H 2 O ? moles? mol H 2 O

Figure 9.2: A map of the procedure used in Example 9.7.

9.7 Examples Aluminum reacts with chlorine gas to form aluminum chloride according to the following reaction: 2Al(s) + 3Cl 2 (g) 2 AlCl 3 (s) In a certain experiment, 10.0 g of aluminum Is reacted with 35.0 g of chlorine gas. What mass of aluminum chloride will be Produced, assuming a complete reaction? What mass of which reactant is left after the reaction? 43.9 g AlCl g Al(s)

9.8 Percent Yield Objectives: To learn to calculate actual yield as a percentage of theoretical yield Actual yield Theoretical Yield x 100 = percent yield

9.8 Percent Yield Suppose 6.85x10 4 g of CO(g) reacts with 8.60 kg of H 2 (g). Calculate the theoretical yield of methanol If 3.57x 10 4 g of CH 3 OH is produced what is the percent yield?

9.8 Percent Yield Urea is formed by the reaction of ammonia and carbon dioxide: 2NH 3 (g) + CO 2 (g) CN 2 H 4 O(s) + H 2 O(l) In a certain experiment g of ammonia is reacted with g of carbon dioxide, and g of urea is produced. Determine the percent yield for this reaction %

The Nuts and Bolts of Chemistry Do Activity on p. 254 Do this in 3 groups Answer the questions in number 8. Turn in the assignment once it’s done. You have 20 minutes.

We are going to a DELI! You are making a sandwich. What do you need? 2 pieces of bread + 3 slices of meat + 1 slice cheese 1 SANDWICH NOW, HOW MUCH WOULD YOU NEED TO FEED THIS CLASS?