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Zumdahl Zumdahl DeCoste CHEMISTRY World of. Chapter 9 Chemical Quantities.

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Presentation on theme: "Zumdahl Zumdahl DeCoste CHEMISTRY World of. Chapter 9 Chemical Quantities."— Presentation transcript:

1 Zumdahl Zumdahl DeCoste CHEMISTRY World of

2 Chapter 9 Chemical Quantities

3 Goals of Chapter 9 Understand molecular and molar mass given in balanced equation Use balanced equation to determine the relationships between moles of reactants and moles of products Relate masses of reactants and products in a chemical reaction Perform mass calculations that involve scientific notation Understand the concept of limiting reactants Recognize the limiting reactant in a reaction Use the limiting reactant to do stoichiometric calculations

4 Information Given by Chemical Equations Reactions are described by equations Give the identities of the reactants & products Show how much of each reactant and product participates in the reaction. Numbers or coefficients enable us to determine how much product we get from a given quantity of reactants

5 Information Given by Chemical Equations Atoms are rearranged in a chemical reaction (not created or destroyed) Must have same number of each type of atom on both sides of equation. Coefficients give relative number of molecules, meaning we can multiply by any number and still have a balanced equation.

6 Table 9.1

7 Combustion of Propane Propane reacts with oxygen to produce heat and the products carbon dioxide and water: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)

8 Interpretation of Equation 1 molecule of C 3 H 8 reacts with 5 molecules of O 2 to give 3 molecules of CO 2 plus 4 molecules of H 2 O 1 mole of C 3 H 8 reacts with 5 moles of O 2 to give 3 moles of CO 2 plus 4 moles of H 2 O

9 Nuts & Bolts of Chemistry Activity 2 Nuts (N) react with 1 bolt (B) to form a nut- bolt molecule 2N + B → N 2 B Note difference between coefficient and subscript Construct nut-bolt molecules Which is limiting reactant? Why?

10 Average mass of bolt = g & average mass of nut = 4.35 g; If you are given about 1500 g of each, answer the following questions: 1. How many bolts are in 1500 g? How many nuts are in 1500 g? 2. Which is limiting reactant? Why? 3. What is largest possible mass of product? How many products can you make? 4. What is mass of leftover reactant?

11 Mole-Mole Relationships 2H 2 O(l) → 2H 2 (g) + O 2 (g) Equation tells us that 2 mol of H 2 O yields 2 mol of H 2 and 1 mol of O 2 I f we decompose 4 mol of water, how many moles of products do we get? 4H 2 O(l) → 4H 2 (g) + 2O 2 (g) If we decompose 5.8 mol of water, how many moles of products do we get? 5.8H 2 O(l) → 5.8H 2 (g) + ?O 2 (g)

12 Mole Ratios: conversion factors based on balanced chemical equations From initial equation: 2 mol H 2 O = 2 mol H 2 = 1 mol O 2 Can use equivalent statement & perform dimensional analysis 5.8 mol H 2 O x 1 mol O 2_ = 2.9 mol O 2 2 mol H 2 O

13 Mass Calculations What mass of oxygen (O 2 ) is required to react with exactly 44.1 g of propane (C 3 H 8 )?

14 Step 1: Write balanced equation C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)

15 44.1 g C 3 H 8 x 1 mol C 3 H 8 = 1.00 mol C 3 H g C 3 H 8 Step 2: Convert grams of propane to moles of propane

16 Step 3: Use coefficients in equation to determine moles of oxygen required 1.00 mol C 3 H 8 x 5 mol O 2 = 5.00 mol O 2 1 mol C 3 H 8

17 Step 4: Use molar mass of O 2 to calculate the grams of oxygen 5.00 mol O 2 x 32.0 g O 2 = 160 g O mol O 2

18 Can perform conversion in on long step: 44.1 g C 3 H 8 x 1 mol C 3 H 8 x 5 mol O 2 x 32.0 g O 2 = 160 g O g C 3 H 8 1 mol C 3 H mol O 2

19 Mass Calculations Using Scientific Notation Step 1: Balance the equation for the reaction Step 2: Convert the masses of reactants or products to moles Step 3: Use the balanced equation to set up the appropriate mole ratio(s) Step 4: Use the mole ratio(s) to calculate the number of moles of the desired product or reactant. Step 5: Convert from moles back to mass

20 Stoichiometry The process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction

21 Mass Calculations: Comparing Two Reactions Antacids are used to neutralize excess hydrochloric acid secreted by the stomach. Which antacid is more effective: baking soda, NaHCO 3, or milk of magnesia, Mg(OH) 2 ? Determine how many moles of stomach acid (HCl) will react with 1.00 g of each acid.

22 The Concept of Limiting Reactants

23 Consider the reaction that forms ammonia: N 2 (g) + 3H 2 (g) → 2NH 3 (g) These gases are mixed in a closed vessel and begin to react

24 Container (#1) of N 2 (g) and H 2 (g).

25 Before and after the reaction.

26 This reaction contained the exact number of molecules to make ammonia molecules with no unreacted molecules left over. Before the reaction, there were 15 H 2 molecules and 5 N 2 molecules which gives the exact ratio to make ammonia, 3:1. This type of mixture is called a stoichiometric mixture – contains the relative amounts of reactants that matches the numbers in balanced equation

27 What happens when the ratio is not the same as in the chemical equation?

28 Container (#2) of N 2 (g) and H 2 (g). (5:8 ratio)

29 Before and after the reaction. (Some N 2 molecules are left over)

30 Limiting Reactant The reactant that runs out first and thus limits the amounts of products that can form H 2 is limiting reactant in previous slide

31 Ammonia, which is used as a fertilizer, is made by combining nitrogen from the air with hydrogen. The hydrogen is produced by reacting methane (CH 4 ) with water. If you have 249 grams of methane, how much hydrogen will be produced and how much water will you need to convert all of the methane to hydrogen?

32 Step 1: Write balanced equation CH 4 (g) + H 2 O(g) → 3H 2 (g) + CO(g)

33 Figure 9.1: A mixture of 5CH 4 and 3H 2 O molecules.

34 Step 2: Convert mass of methane to moles 249 g CH 4 x 1 mol CH 4 = 15.5 mol CH g CH 4

35 Step 3: Determine moles of H 2 O needed 15.5 mol CH 4 x 1 mol H 2 O = 15.5 mol H 2 O 1 mol CH 4

36 Step 4: Determine mass of water 15.5 mol H 2 O x g H 2 O = 279 g H 2 O 1 mol H 2 O

37 Reacting 279 grams of water with 249 grams of methane will cause both reactants to “run out” at the same time. If 300 grams of water is reacted with 249 grams of methane, the methane will “run out” first = limiting reactant (it limits the reaction)

38 Figure 9.2: A map of the procedure used in Example 9.7. (see for determining limiting reactant)

39 Steps for solving stoichiometric problems involving limiting reactants: Step 1: Write and balance chemical equation Step 2: Convert known masses to moles Step 3: Using number of moles of reactants and mole ratios, determine limiting reactant Step 4: Use amount of limiting reactant and mole ratios to calculate number of moles of product Step 5: Convert from moles to mass

40 Theoretical Yield Calculated yield from chemical reaction (amount of product form mole ratio calculations) Maximum amount that can be produced Amount predicted is seldom obtained Side reactions occur Actual yield: amount of product actually obtained

41 Actual Yield x 100% = percent Theoretical Yield yield Percent Yield


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