CHEM 101 1nd Edition Lecture 15Outline of Last Lecture I. Assigning Oxidation StatesII. Oxidation State ReactionsOutline of Current Lecture I. StoichiometryII. Stoichiometric CalculationsIII. Percent YieldIV. Other CalculationsCurrent LectureI. Stoichiometrya. A balanced reaction shows the quantitative relationship between reactants and products in a chemical reactionb. Fe2O3 (s) + 3 CO (g) ® 2 Fe (s) + 3 CO2 (g)i. Initial 1 mol 3 mol 0 mol 0 molii. Change -1 mol -3 mol + 2 mol +3 moliii. Final 0 mol 0 mol 2 mol 3 molc. The stoichiometric coefficients do not tell us how much we haved. They tell us the ratios in which reactants are lost and products are formedII. Stoichiometric Calculationsa. General Plani. Mass Reactant --> Moles Reactant --(Stoichiometric Factor)--> Moles Product --> Mas Productb. Example:i. If 454 g of NH4MO3 decomposes. How much N2O and H2O can be formed?In other words, what is the theoretical yield of products?1. Step 1a. Write the balanced equationb. NH4NO3 --> N2O + H2O2. Step 2a. Convert mass reactant, 454 g --> Molesb. Formula weightThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.i. 2N = 2 x 14.0067 = 28.0134ii. 4H = 4 x 1.0079 = 4.0316iii. 3O = 3 x 15.9994 = 46.9982iv. NH4NO3 80.0435 g/molc. 454 g x 1 mol / 80.0435 g = 5.86 mol NH4NO33. Step 3a. Convert moles reactant --> moles productb. 5.86 mol NH4NO3 x (2 mol H2O produced / 1 mol NH4NO3 used)c. = 11.4 mol H2O produced4. Step 4a. Convert moles product --> mass producti. Called the Theoretical Yieldb. 11.4 mol H2O x (18.02 g / 1 mol) = 204 g H2O5. Step 5: How much N2O is formed?a. We can go through all the steps we just did orb. You can remember that the total mass or reactants must equal the total mass of product. c. So if all 454 g of NH4NO3 are consumed, and we know we have 204 g H2O we therefore have 350 g NO3.i. 454 NH4NO3 = ? g N2O + 204 g H2Od. But we have to make sure 100% of the reactants are used6. NOTE: Normally these would be done in one continues calculationc. Amounts TableCompound NH4NO3N2O H2OInitial (g) 454 g 0 0Initial (mol) 5.68 mol 0 0Change (mol) -5.68 +5.68 +2(5.68)Final (mol) 0 5.68 11.4Final (g) 0 250. 204i.ii. Mass is conserved moles are notIII. Percent Yield from Chemical Reactionsa. Many reaction in real life do not go to completioni. Reactants are not completely transformed into the desired productsb. At least 2 reason for thisi. Reactants did not react entirely to give the productsii. Reacts did react 100% but formed "undesirable" products by "see reactions" as well as the wanted productsc. Theoretical Yield: R1 + R2--> 100% completiond. Percent Yield: % Yield = (Actual Yield / Theoretical Yield) x 100%e. Step 6i. Calculate % Yieldii. % Yield = (131 g / 250. g) x 100% = 52.4%IV. Other Examplesi. What is the mass of CO required to react with 62.4 g Fe2O31. Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) + 3 CO2 (g)2. Plan:a. g Fe2O3 --> mol Fe2O3 --> mol CO --> g CO3. 61.4 g Fe2O3 (1 mol Fe2O3 / 159.7 g FeO2)(3 mol CO / 1 mol Fe2O3)(28.0 g CO / 1 mol CO) = 32.2 gii. What is the mass of Fe2O3 reacted with excess CO if the CO2 produced by the reaction has a mass of 5.86?1. Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) 3 CO2 (g)2. Plan:a. g CO2 --> mol CO2 --. Mole Fe2O3 --> g Fe2O33. 5.86 g CO2 (1 mol CO2 / 44 g CO2)(1 mol Fe2O3 / 3 mol CO2)(159.7 g Fe2O3 / 1 mol Fe2O3) = 7.09 giii. How many grams of Fe2O3 are required to produce 10.0 g of Fe assuming a percent yield of 85%?1. Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) + 3 CO2 (g)2. Plan:a. Calculate theoretical yield from mass of Fe and % yieldb. Using theoretical yield to calculate mass of Fe2O3i. % Yield = (actual / theoretical) x 1003. 11.8 g Fe (1 mol Fe / 55.85 g Fe)(1 mol Fe2O3 / 2 mol Fe)(159.7 g Fe2O3 / 1 mol Fe2O3) = 16.9