BIOM 121 1nd EditionExam # 1 Study Guide Lectures: 1 - 12Lecture 15 (September 30)I. 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 weighti. 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 gFe2O3 / 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 gLecture 16 (October 2) I. Limiting Reagenta. Limiting Reagenti. Reactant that is the first to be completely consumed in a reactionb. Excess Reagenti. Any reactant that remains once the limiting reagent has been completely consumedc. Examplesi. Mix 5.40 g Al with 8.10 g Cl2. What mass of Al2Cl6 can form? 1. Step 1a. Compare actual mole ratio of reactant to theoretical mole ratiob. 2 Al + 3 Cl2 --> Al2Cl6i. Reactant must be in the mole ratio1. (mol Cl2 / mol Al) = 3/2ii. If (mol Cl2 / mol Al) > 3/2 then Cl2 is in excess and Alis the limiting reagentiii. If (mol Cl2 / mol Al) < 3/2 then Cl2 is the limiting reagent. 2. Step 2a. Calculate moles of each reactantb. We have 5.40 g Al and 8.10 g Cl2i. 5.40 g Al x (1 mol/27.0) = .200 mol Alii. 8.10 g Cl2 x (1 mol / 70.9 g) = .114 mol Cl2c. Find mole ratio of resultsi. (Mol Cl2/ mol Al) = (0.114 mol / 0.200 mol) = 0.57 1. This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio.2. Therefore the limiting reactant is Cl.3. Step 3, Calculationsa. Step 1: Calculate mols of Al2Cl6 expected based on limiting reactanti. .114 mol Cl2 x (1 mol Al2Cl6 / 3 mol Cl2) = .0380 molAl2Cl6b. Step 2: Calculate mass of Al2Cl6 expected bas on limiting reagenti. .0380 mol Al2Cl6 x (266.4 g Al2Cl6 / mol) = 10.1 g Al2Cl64. How much of which reactant will remain when the reaction is complete?a. Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much?b. First find how much Al was requiredc. Then find how much Al is in excessd. Calculation Excess Ali. 2 Al + 3 Cl2  Productsii. .114 mol Cl2 x (2 mol Al / 3 mol Cl2) = .0760 mol Al requirediii. Excess Al = Al available – Al required1. .200 mol - .0760 mol2. .124 mol Al in excessII. Concentration and Molaritya. Concentrationsi. The amount of solute present in a given quantity of solvent or solutionb. Molarityi. aka Molar Concentration, moles of solute per liter of solutionii. M = (moles of solute / volume of solution in liters)c. Using Molarityi. What mass of oxalic acid is required to make 250. mL of a .0500 M solution?1. Because the Concentration (M) = moles/volume = mol/Va. This means that moles = M x V2. Step 1: Calculate moles of acid requireda. (.0500 mol/L)(.250 L) = .0125 mol3. Step 2: Calculate mass of acid requireda. (.0125 mol)(90.00 g/mol) = 1.13 gLecture 17 (October 4) I. pH: Potential Hydrogena. Acidity is measured in terms of [H3O+]i. Typically ranges from 1 to 1 x 10-14 Mb. pH = - log [H3O+]i. Typically ranges from 0 to 14ii. pH < 7 is acidiciii. pH > 7 is basicII. pH and [H+]a. Because pH = - log [H+] then, log [H+] = - pHi. Take antilog and get: [H+] = 10-pHb. Neutralization:i. An acid and a base react to form water and a dissolved saltii. Equivalence pout is when there is an …