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Zumdahl • Zumdahl • DeCoste

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

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Zumdahl • Zumdahl • DeCoste

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  1. CHEMISTRY Zumdahl • Zumdahl • DeCoste 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. Propane reacts with oxygen to produce heat and the products carbon dioxide and water: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Combustion of Propane

  8. Interpretation of Equation • 1 molecule of C3H8 reacts with 5 molecules of O2 to give 3 molecules of CO2 plus 4 molecules of H2O • 1 mole of C3H8 reacts with 5 moles of O2 to give 3 moles of CO2 plus 4 moles of H2O

  9. Nuts & Bolts of Chemistry Activity • 2 Nuts (N) react with 1 bolt (B) to form a nut-bolt molecule • 2N + B → N2B • Note difference between coefficient and subscript • Construct nut-bolt molecules • Which is limiting reactant? Why?

  10. Average mass of bolt = 10.64 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. 2H2O(l) → 2H2(g) + O2(g) Equation tells us that 2 mol of H2O yields 2 mol of H2 and 1 mol of O2 If we decompose 4 mol of water, how many moles of products do we get? 4H2O(l) → 4H2(g) + 2O2(g) If we decompose 5.8 mol of water, how many moles of products do we get? 5.8H2O(l) → 5.8H2(g) + ?O2(g) Mole-Mole Relationships

  12. From initial equation: 2 mol H2O = 2 mol H2 = 1 mol O2 Can use equivalent statement & perform dimensional analysis 5.8 mol H2O x 1 mol O2_= 2.9 mol O2 2 mol H2O Mole Ratios: conversion factors based on balanced chemical equations

  13. What mass of oxygen (O2) is required to react with exactly 44.1 g of propane (C3H8)? Mass Calculations

  14. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Step 1: Write balanced equation

  15. 44.1 g C3H8 x 1 mol C3H8 = 1.00 mol C3H8 44.09 g C3H8 Step 2: Convert grams of propane to moles of propane

  16. 1.00 mol C3H8 x 5 mol O2= 5.00 mol O2 1 mol C3H8 Step 3: Use coefficients in equation to determine moles of oxygen required

  17. 5.00 mol O2x 32.0 g O2= 160 g O2 1.00 molO2 Step 4: Use molar mass of O2 to calculate the grams of oxygen

  18. 44.1 g C3H8 x 1 mol C3H8 x 5 mol O2x32.0 g O2= 160 g O2 44.09 g C3H8 1 mol C3H8 1.00 molO2 Can perform conversion in on long step:

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

  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, NaHCO3, 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. N2(g) + 3H2(g) → 2NH3(g) These gases are mixed in a closed vessel and begin to react Consider the reaction that forms ammonia:

  24. Container (#1) of N2(g) and H2(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 H2 molecules and 5 N2 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 N2(g) and H2(g). (5:8 ratio)

  29. Before and after the reaction. (Some N2 molecules are left over)

  30. Limiting Reactant • The reactant that runs out first and thus limits the amounts of products that can form • H2 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 (CH4) 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. CH4(g) + H2O(g) → 3H2(g) + CO(g) Step 1: Write balanced equation

  33. Figure 9.1: A mixture of 5CH4 and 3H2O molecules.

  34. 249 g CH4 x 1 mol CH4 = 15.5 mol CH4 16.04 g CH4 Step 2: Convert mass of methane to moles

  35. 15.5 mol CH4 x 1 mol H2O = 15.5 mol H2O 1 mol CH4 Step 3: Determine moles of H2O needed

  36. 15.5 mol H2O x 18.02 g H2O = 279 g H2O 1 mol H2O Step 4: Determine mass of water

  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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