CHEM 101 1nd Edition Lecture 16Outline of Last Lecture I. StoichiometryII. Stoichiometric CalculationsIII. Percent YieldIV. Other CalculationsOutline of Current Lecture I. Limiting ReagentII. Concentration and MolarityCurrent LectureI. 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 Cl2These 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.c. 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 mol Al2Cl6b. 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
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