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Chapter 4

Chapter 4. Chemical Reactions. Changing Matter. Chemistry is defined as the study of matter and changes it undergoes. In chemistry, we recognize two different types of ways that matter can change. Physical Change Chemical Change. Physical Changes.

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Chapter 4

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  1. Chapter 4 Chemical Reactions

  2. Changing Matter • Chemistry is defined as the study of matter and changes it undergoes. In chemistry, we recognize two different types of ways that matter can change. • Physical Change • Chemical Change

  3. Physical Changes • A physical change is a change in only the appearance of matter. The matter does not change identity, only its form. • Boiling water is a physical change. The water changes from a liquid to a gas, so it changes its form or appearance. But it remains water throughout the entire process. • Other examples are freezing a liquid, pressing plastic into a new shape, or breaking an object.

  4. Chemical Changes • A chemical change is a change in a substance’s identity. During a chemical change, the atoms in a compound or molecule rearrange to form a new compound or molecule. • Chemical changes are also known as chemical reactions. They may or may not also involve a physical change. • Examples inclued • An iron nail turning to rust • Burning propane gas • Rotting or decomposition of organic materials.

  5. Chemical Equations • A chemical equation is a recipe for a chemical reaction. A chemical equation contains all the same type of information as a recipe. • What four things are in a recipe? • Reactants • Products • Coefficients • Arrow

  6. Chemical Reactions “reactants”“products” starting material produced by rxn.

  7. Chemical Equations CH4+ 2O2 CO2 + 2H2O “coefficients”“subscripts” coefficientscan be changed to achieve mass balance subscripts are never changed to balance an equation

  8. Chemical Equations H2O + 3 NO2 2 HNO3 + NO • Identify: • Reactants • Products • Coefficients • Arrow

  9. Molar Mass • The molar mass is the mass of one mole of anything. For example, the mass of a mole of neon atoms is known as its molar mass. The mass of a mole of butane molecules is known as its molar mass. Also known as formula weight • Molecular weight is used only for molecules. • For elements, the atomic mass (on the periodic table) is equal to its molar mass in grams.

  10. Molar Mass • Molar mass can be used as a conversion factor • The atomic mass of potassium is 39.10. 1 mole of potassium weighs 39.10 grams. 1 mole = 39.10 g • What is the molar mass of neon? • What is the molar mass of oxygen?

  11. Molar Mass of Compounds • For compounds, the molar mass is the sum of the molar masses of each individual atom in the compound. • Problem 4.1 For: KClO3 K 39.098 amu * 1 = 39.098 amu Cl 35.453 amu * 1 = 35.453 amu O 15.999 amu * 3 = 47.997 amu Formula Weight of KClO3 = 122.548 amu

  12. The Mole • The mole is exactly like the dozen. It is a way of counting a large number of items. • A dozen represents the number 12, 12 eggs can be called a dozen eggs • The mole represents the number 6.02 x 1023, 6.02 x 1023 eggs can be called a mole of eggs.

  13. The Mole • We use the mole to quickly count atoms and molecules, just like we use the dozen to quickly count eggs. • In the same way that eggs are packaged and sold by the dozen, we often work with atoms and molecules by the mole.

  14. Avogadro's Number • 6.02214199 x 1023 is known as Avogadro’s number. • Avogadro’s number can be used as a conversion factor: 1 mole = 6.02 x 1023.

  15. Practice Problems • How many molecules of aspirin (C9H8O4) are in 1.75 mol? • How many atoms of oxygen are in 2.1 mol aspirin? • How many atoms of hydrogen are in 1.7 mol aspirin? • How many atoms of carbon are in 3.2 mol aspirin?

  16. Molar Mass as a Conversion • in 25.0 g of boron? • in 342.3 g of sucrose? • in 10.0 g of sucrose?

  17. Molar Mass as a Conversion Factor • Problem 4.2 p 112 • Problem 4.3 p 112 • Problem 4.4 p 113 • Problem 4.5 p 113 • Problem 4.6 p 114

  18. Conservation of Mass inChemical reactions • Atoms cannot be created or destroyed by ordinary chemical reactions. Therefore, all atoms which are reacting in a chemical reaction must also show up as a product of that reaction. • When there is an equal number of each type of atom on both sides of the arrow in a chemical equation, it is said to be balanced.

  19. Balancing Chemical Equations • Write formula for each reactant and product on the correct side of the “reaction arrow” • Count atoms of each element on both sides of arrow • Start with the compound which has the most complex formula • Add coefficients to chemical formulas to balance numbers of each atom • Trial and error begins... May add subscripts of (g), (s), (l), (aq) or (ppt) to indicate the physical state of the participants.

  20. Balancing Chemical Equations Examples: ___Al + ___Cl2 ___AlCl3 ___Ca + ___H2O ___Ca(OH)2 + ___H2 ___HCl + ___Al(OH)3 ___AlCl3 + ___H2O ___Ba(OH)2 + ___H3PO4 ___Ba3(PO4)2 + ___H2O Combustion Example: ___C5H8O3 + ___O2 ___CO2 + ___ H2O

  21. Practice Problems (p 216 & 217) • Problem 4.7 • Problem 4.8 • Problem 4.9

  22. Types of Reactions • 4 different types of Reactions: • Combination: A + B AB • Decomposition: AB  A + B • Single Replacement: A + BC  AC + B • Double Replacement: AB + CD  AD + BC

  23. Mole Relationships in Chemical Equations • N2(g)+ 3 H2 (g)  2NH3 (g) • What are the mole to mole relationships? • How many moles of hydrogen are needed to react with 1.3 moles of nitrogen? • How many moles of nitrogen reacted in 0.60 moles of ammonia is produced? • How many moles of ammonia are produced when 1.4 moles of hydrogen react?

  24. Mass Calculations for Reactions • N2(g)+ 3 H2 (g)  2NH3 (g) • If you have 1.80 moles of hydrogen, how many grams of ammonia can be produced? • How many grams of hydrogen are needed to react with 2.80 g of nitrogen? • How many grams of ammonia can be produced from 12.0 g of hydrogen?

  25. Mass Calculations for Chemical Reactions • Problem 4.10 p 119 • Problem 4.11 p 120 • Problem 4.12 p 121

  26. Limiting Reactants and Reagents • The limiting reactant is the reactant in a chemical reaction which limits the amounts of products that can be formed • The limiting reagent in a chemical reaction is present in insufficient quantity to consume the other reactant(s)

  27. Limiting Reagent

  28. Calculating Limiting Reagents 1) calculate moles (or mass) of product formed by complete reaction of each reactant. 2) the reactant that yields the least product is the limiting reagent. 3) the theoretical yield for a reaction is the maximum amount of product that could be generated by complete consumption of the limiting reagent. Problem 4.13 p 122

  29. Limiting Reactants • Iron (III) oxide reacts with carbon monoxide to produce iron and carbon dioxide. • What is the limiting reactant if 43.2 g iron (III) oxide reacts with 24.3 g carbon monoxide? • What is the theoretical yield?

  30. Percent Yield • Theoretical yield– the maximum amount of product expected in a chemical reaction. • In reality, most chemical reactions produce less product than the theoretical yield predicts. The actual amount of product obtained from a chemical reaction is the actual yield. The actual yield is only determined by doing a reaction and weighing the amount of product. • The percent yield is: (actual yield) / (theoretical yield) x 100.

  31. Percent Yield • Carbon disulfide and carbon monoxide are produced by the reaction of carbon and sulfur dioxide: • What is the percent yield for the reaction if 40.0 g of carbon produces 36.0 g of carbon disulfide?? • What is the percent yield for the reaction if 32.0 g of sulfur dioxide produces 12.0 g of carbon disulfide? • What is the percent yield for the reaction if 40.0 g of carbon reacts with 12.0 g of sulfur dioxide to form 37 g of carbon disulfide?

  32. Calculating Percent Yield When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction: 2CH4 + 2NH3 + 3O2 2HCN + 6H2O What is the percent yield of HCN in this reaction? How many grams of NH3 remain?

  33. Calculating Limiting Reagent 66.6 g of O2 = 2.08 mol O2 27.8 g of NH3 = 1.63 mol NH3 25.1 g of CH4 = 1.56 mol CH4 Which reactant is limiting? 2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN 1.63 mol NH3 can yield 1.63 mol (or 44.1 g) HCN 1.56 mol CH4 can yield 1.56 mol (or 42.2 g) HCN O2 is the limiting reagent.

  34. O2 is the limiting reagent; thus, the theoretical yield is based on 100% consumption of O2. 2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN % yield = actual yield * 100 theoretical yield % yield = 36.4 g HCN = 97.1% * 100 37.5 g HCN

  35. 2CH4 + 2NH3 + 3O2 2HCN + 6H2O How many grams of NH3 remain? 36.4 g (or 1.35 mol) of HCN gas is produced Since the reaction stoichiometry is 1:1, 1.35 mol of NH3 is consumed: (1.63 mol NH3 initially) –(1.35 mol NH3 consumed) = 0.28 mol NH3 remaining 0.28 mol NH3 * (17.03 NH3 g/mol NH3) = 4.8 g NH3 remain

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