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CHAPTER 3 Chemical Equations & 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 2 Equations Consider a simple equation: 23 ? The question is: what happens to 2 if we multiply it by 3? 23 6 The answer: it is converted to 6 In chemistry, we try to answer similar problems using chemical equations 3 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 4 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 unbalanced equation 5 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 reactants products 6 Chemical Equations CH4 + 2O2 CO2 + 2H2O 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 7 Balancing Chemical Equations Fe + O2 Fe3O4 Al + H2SO4 Al2(SO4)3 + H2 C6H12O6 + O2 CO2 + H2O 8 Quantitative Aspects Let’s go back to 23 6 If we multiply both sides of the equation by some number, they still will be equal 8 (2 3) 8 6 We can treat chemical equations in the same way 9 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: CH4 + 2O2 CO2 + 2H2O 1 molecule 2 molecules 1 molecule 8 molecules 16 molecules 8 molecules 2 molecules x8 16 molecules 10 Example 1 How many O2 molecules are required to react with 81 atoms of Fe? Fe + O2 Fe3O4 11 Quantitative Aspects Let’s multiply the equation by Avogadro’s number: CH4 + 2O2 CO2 + 2H2O 1 molecule 2 molecules 1 molecule 2 molecules 6.022x1023 molecules 2x(6.022x1023 molecules) 1 mole = 2 moles 6.022x1023 molecules = = = 1 mole 2x(6.022x1023 molecules) 2 moles In the same way as we talk about chemical equations in terms of molecules, we can consider them in terms of moles 12 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 13 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 14 Example 3 What mass of CO is required to react with 146 g of iron(III) oxide? Fe2O3 + CO Fe + CO2 15 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 16 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 17 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 18 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, 132.03g 159.69g 8.65 g ?g ?g 159.69g x 8.65 g 10.5 g Fe2O3 132.03g 19 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 20 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? CS2 O2 CO2 SO2 21 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 1) Balance the equation and calculate molar masses of all substances (since all of them are included in the formulation of the problem): C M r: 12.01 g/mol + 2Cl2 70.91 g/mol CCl4 153.82 g/mol 22 Example 7 (continued) 2) Calculate the amount of reactants in moles by dividing the mass of each by its molar mass: C m: Mr : 10.0 g 12.01 g/mol 0.833 mol + 2Cl2 CCl4 100.0 g 70.91 g/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 23 Example 7 (continued) 3) Now, knowing the amount of the limiting reactant, we can easily calculate the amount of product. According to the equation, C + 2Cl2 CCl4 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. m (CCl4) = #moles(CCl4) x Mr(CCl4) = = 0.705 mol x 153.82 g/mol = 108 g 24 Example 7 (continued) 4) 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 25 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 26 Percent Yields from Reactions Actual yield is the amount of a specified pure product actually obtained from a given reaction Percent yield: actual yield % yield= 100% theoretical 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 27 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 1) We start by calculating the molar masses of C2H5OH and CH3COOC2H5: Mr(C2H5OH) = 46.08 g/mol Mr(CH3COOC2H5) = 88.12 g/mol 2) The amount of C2H5OH in moles is 10.0 g # moles(C2H5OH) = 0.217 mol 46.08g/mol 28 Example 8 (continued) C2H5OH + CH3COOH CH3COOC2H5 + H2O 3) 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 14.8 g percent yield= 100% 77.5% 19.1 g 29 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 1) Calculate the number of moles of each reactant to determine which reactant is the limiting one: 10.6 g m(Fe) # moles(Fe) = 0.190 mol Mr (Fe) 55.85g/mol 25 g m(Br2 ) # moles(Br2 ) = 0.156 mol Mr (Br2 ) 159.81g/mol 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. 30 Example 9 (continued) 2) 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 3) 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 31 Example 9 (continued) 4) Now we can write the following relationship: 479.43 g Br2 gives 591.12 g FeBr3 25 g Br2 gives ? g FeBr3, Therefore, 479.43g 591.12g 25 g ?g ?g 591.12g x 25 g 30.8 g FeBr3 479.43g this is theoretical yield 18 g percent yield= 100% 58% 30.8 g 32 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 33 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? 1) The balanced sequential equations are: 2C10H22 + 31O2 20CO2 + 22H2O 2NaOH + CO2 Na2CO3 + H2O 2) We calculate the amount of C10H22 in moles: # moles(C10H22 ) = m(C10H22 ) 13 g 0.091 mol Mr (C10H22 ) 142.28g/mol 34 Example 10 (continued) 2C10H22 + 31O2 20CO2 + 22H2O 2NaOH + CO2 Na2CO3 + H2O 3) 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 4) 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. 35 Example 10 (continued) 5) 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 36 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%) 37 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 38