1 / 38

CHAPTER 3

CHAPTER 3. Chemical E quations & Reaction Stoichiometry. Objectives. Understand how to write chemical equations Perform calculations based on chemical equations Calculate percent yields from chemical reactions Understand the concept of sequential reactions. Equations.

talli
Download Presentation

CHAPTER 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 3 Chemical Equations & Reaction Stoichiometry

  2. Objectives • Understand how to write chemical equations • Perform calculations based on chemical equations • Calculate percent yields from chemical reactions • Understand the concept of sequential reactions

  3. Equations • Consider a simple equation: • The question is: what happens to 2 if we multiply it by 3? • The answer: it is converted to 6 • In chemistry, we try to answer similar problems using chemical equations

  4. Chemical Equations CH4 + O2 ? • We ask: what happens to methane when it reacts with oxygen (burns)? • We know the answer from the chemical experiment CH4 + O2 CO2 + H2O • The only thing we have to check is the Law of Conservation of Matter

  5. unbalanced equation Chemical Equations • The Law of Conservation of Matter: in any physical or chemical change the total mass of matter remains constant which means the number of atoms of each element involved remains unchanged CH4 + O2 CO2 + H2O

  6. reactants products Chemical Equations • We need to: (1) balance the equation CH4 + O2 CO2 + H2O (2) make sure that we have the same number of atoms for each element on the left and on the right side CH4 + 2O2 CO2 + 2H2O

  7. Chemical Equations • Symbolic representation of a chemical reaction that shows: • reactants on the left side • products on the right side • relative amounts of each using stoichiometric coefficients CH4 + 2O2 CO2 + 2H2O

  8. Balancing Chemical Equations • Fe + O2 Fe3O4 • Al + H2SO4 Al2(SO4)3 + H2 • C6H12O6 + O2 CO2 + H2O

  9. Quantitative Aspects • Let’s go back to • If we multiply both sides of the equation by some number, they still will be equal • We can treat chemical equations in the same way

  10. CH4 + 2O2 CO2 + 2H2O 1 molecule 2 molecules 1 molecule 2 molecules x 8 Quantitative Aspects • What happens in the reaction between methane and oxygen numerically? CH4 + 2O2 CO2 + 2H2O 1 molecule 2 molecules 1 molecule 2 molecules • We can multiply both left and right sides by the same number – they will remain equal: 8 molecules 16 molecules 8 molecules 16 molecules

  11. Example 1 • How many O2 molecules are required to react with 81 atoms of Fe? Fe + O2 Fe3O4

  12. 6.022x1023 molecules 2x(6.022x1023 molecules) 6.022x1023 molecules 2x(6.022x1023 molecules) = = = = 1 mole 2 moles 1 mole 2 moles Quantitative Aspects • Let’s multiply the equation by Avogadro’s number: CH4 + 2O2 CO2 + 2H2O 1 molecule 2 molecules 1 molecule 2 molecules • In the same way as we talk about chemical equations in terms of molecules, we can consider them in terms of moles

  13. Example 2 • How many moles of H2 is produced when 70 moles of aluminum react with excess sulfuric acid? Al + H2SO4 Al2(SO4)3 + H2

  14. Quantitative Aspects • Since we know the molar mass of each substance we can also establish the mass relationships: CH4 + 2O2 CO2 + 2H2O 1 mole 2 moles 1 mole 2 moles x 16.0 g/mol 32.0 g/mol 44.0 g/mol 18.0 g/mol 16.0 g + 64.0 g = 44.0 g + 36.0 g • The total mass of products should be the same as the total mass of reactants

  15. Example 3 • What mass of CO is required to react with 146 g of iron(III) oxide? Fe2O3 + CO  Fe + CO2

  16. Example 4 • What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron(III) oxide with excess carbon monoxide? Fe2O3 + 3CO  2Fe + 3CO2 • We should be concerned only with iron(III) oxide and carbon dioxide since carbon monoxide is in excess. Therefore, we calculate the number of moles of CO2 which can be produced from 0.540 mole of Fe2O3: • 1 mole of Fe2O3 produces 3 moles of CO2, hence • 0.540 mole of Fe2O3 will produce (0.540 x 3) mole = 1.62 mole of CO2

  17. Example 4 (continued) • Now we can calculate the mass of CO2 produced: • mass(CO2) = #moles(CO2) x molar mass (CO2) mass(CO2) = 1.62 mole x 44.0 g/mol = 71.28 g

  18. Example 5 • What mass of iron(III) oxide reacted with excess carbon monoxide if carbon dioxide produced by the reaction had a mass of 8.65 grams? Fe2O3 + 3CO  2Fe + 3CO2 • We should be concerned only with iron(III) oxide and carbon dioxide. First, we calculate their molar masses: • Mr(Fe2O3) = 159.69 g/mol • Mr(CO2) = 44.01 g/mol

  19. Example 5 • From the equation we see that 1 mole of Fe2O3 affords 3 moles of CO2. We can convert these quantities to grams: • m (Fe2O3) = 1 mole x 159.69 g/mole = 159.69 g • m (CO2) = 3 mole x 44.01 g/mole = 132.03 g • Now we can write the following relationship: • 132.03 g CO2 is obtained from 159.69 g Fe2O3 8.65 g CO2 is obtained from ? g Fe2O3 • Therefore,

  20. Limiting Reactant Concept • Example: A box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box? • 87 bolts  • 110 washers  • 99 nuts 

  21. Example 6 • What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

  22. Example 7 • Calculate the mass of carbon tetrachloride which can be produced by the reaction of 10.0 g of carbon with 100.0 g of chlorine. Determine the mass of excess reagent left unreacted. C + Cl2  CCl4 • Balance the equation and calculate molar masses of all substances (since all of them are included in the formulation of the problem): C + 2Cl2  CCl4 Mr: 12.01 g/mol 70.91 g/mol 153.82 g/mol

  23. Example 7 (continued) • Calculate the amount of reactants in moles by dividing the mass of each by its molar mass: C + 2Cl2  CCl4 m : 10.0 g 100.0 g Mr: 12.01 g/mol 70.91 g/mol 0.833 mol 1.41 mol It follows from the equation that 1 mole of C reacts with 2 moles of Cl2, therefore 0.833 mole of C should react with 0.833 x 2 = 1.67 mole of Cl2. The amount of Cl2 participating in the reaction is smaller (1.41 mol) which means that Cl2 is the limiting reactant

  24. Example 7 (continued) • Now, knowing the amount of the limiting reactant, we can easily calculate the amount of product. According to the equation, from 2 moles of Cl2 we can obtain 1 mole of CCl4. Hence, from 1.41 mol of Cl2 we can obtain 1.41/2 = 0.705 mol of CCl4. C + 2Cl2  CCl4 m (CCl4) = #moles(CCl4) x Mr(CCl4) = = 0.705 mol x 153.82 g/mol = 108 g

  25. Example 7 (continued) • According to the equation, the number of moles of carbon consumed in the reaction equals to the number of moles of CCl4 formed: C + 2Cl2  CCl4 #moles (C) = 0.705 mol m (C) = #moles(C) x Mr(C) = 0.705 mol x 12.01 g/mol = 8.47 g This means that the amount of carbon that didn’t react is m (C-unreacted) = 10.0 g – 8.47 g = 1.53 g

  26. Percent Yields from Reactions • Theoretical yield is calculated by assuming that the reaction goes to completion • It is the maximum yield possible for the given reaction BUT • in many reactions the reactants are not completely converted to products • a particular set of reactants may undergo two or more reactions simultaneously • sometimes it is difficult to separate the desired product from other products in the reaction mixture

  27. Percent Yields from Reactions • Actual yield is the amount of a specified pure product actually obtained from a given reaction • Percent yield: • There are many reactions that do not give the 100% yield • When calculating the percent yield, always make sure that the actual and theoretical yields are expressed in the same units

  28. Example 8 • A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield? C2H5OH + CH3COOH  CH3COOC2H5 + H2O • We start by calculating the molar masses of C2H5OH and CH3COOC2H5: Mr(C2H5OH) = 46.08 g/mol Mr(CH3COOC2H5) = 88.12 g/mol • The amount of C2H5OH in moles is

  29. Example 8 (continued) C2H5OH + CH3COOH  CH3COOC2H5 + H2O • According to the equation, from 1 mole of C2H5OH we can obtain 1 mole of CH3COOC2H5, therefore from 0.217 mol of C2H5OH we can obtain 0.217 mol of CH3COOC2H5, or theoretical yield = 0.217 mol x 88.12 g/mol = 19.1 g The actual yield, however, is 14.8 g and

  30. Example 9 • 10.6 g of Fe reacts with 25 g of Br2 to form 18 g of FeBr3. What is the percent yield? 2Fe + 3Br2  2FeBr3 • Calculate the number of moles of each reactant to determine which reactant is the limiting one: According to the equation, we need 3 moles of Br2 per each 2 mole of Fe. The calculation shows that we have more moles of Fe than of Br2. Obviously, Br2 is the limiting reagent.

  31. Example 9 (continued) • In this case we don’t even need to perform any further calculations in order to see that Br2 is the limiting reactant. Here’s why: According to the equation, we need 3 moles of Br2 per each 2 mole of Fe. The calculation performed in step (1) shows that we have more moles of Fe than of Br2. Obviously, Br2 is the limiting reactant. 2Fe + 3Br2  2FeBr3 • From the equation we see that each 3 moles of Br2 result in 2 moles of FeBr3. We can convert these quantities to grams (multiplying by the molar mass): • m (Br2) = 3 mole x 159.81 g/mole = 479.43 g • m (FeBr3) = 2 mole x 295.56 g/mole = 591.12 g

  32. Example 9 (continued) • Now we can write the following relationship: 479.43 g Br2 gives 591.12 g FeBr3 25 g Br2 gives ? g FeBr3, Therefore, this is theoretical yield

  33. Sequential Reactions • A set of reactions required to convert starting materials into the desired product • The amount of the desired product from each reaction is taken as the starting material for the next reaction. magnesium perchlorate sodium hydroxide

  34. Example 10 • 13 g of decane, C10H22, is burned in a C-H combustion train. The CO2 gas formed reacts with sodium hydroxide, NaOH, and is converted into sodium carbonate, Na2CO3. What mass of Na2CO3 will be formed if all reactions proceed to completion? • The balanced sequential equations are: 2C10H22 + 31O2  20CO2 + 22H2O 2NaOH + CO2  Na2CO3 + H2O • We calculate the amount of C10H22 in moles:

  35. Example 10 (continued) 2C10H22 + 31O2  20CO2 + 22H2O 2NaOH + CO2  Na2CO3 + H2O • Now, looking at the first equation, we can see that when 2 moles of C10H22 are burned, 20 moles of CO2 are formed, that is to find the number of moles of CO2 formed we need to multiply the given number of moles of C10H22 by 10: #moles(CO2) = 0.091 mol x 10 = 0.91 mol • From the second equation it follows that every mole of CO2 produces 1 mole of Na2CO3, that is the amount of Na2CO3 obtained from 0.91 mol of CO2 will also be equal to 0.91 mol.

  36. Example 10 (continued) • Finally, knowing the number of moles of Na2CO3 we can calculate its mass: m(Na2CO3) = #moles (Na2CO3) x Mr(Na2CO3) m(Na2CO3) = 0.91 mol x 105.99 g/mol = 96 g

  37. Example 11 • What mass of (NH4)3PO4 can be produced in the result of the following reactions if we start with 10 moles of N2 and excess hydrogen? N2 + H2 NH3 (yield = 44%) NH3 + H3PO4 (NH4)3PO4 (yield = 95%)

  38. Reading Assignment • Read Chapter 3 • Learn Key Terms (p. 112) • Take a look at Lecture 5 notes (will be posted on the web not later than Monday morning) • If you have time, read Chapter 4 • Homework #1 due by 9/13

More Related