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University of Ottawa
Kenneth Campbell

Relating the Numbers of Moles of Reactant and Products Problem: How many moles of CO are produ2ed in the combustion of 2.72 mol of triethylene glycol, C 6 14, 4n an excess of O ? 2 An excess of oxygen gas means that triethylene glycol is the limiting reactant. Balance the equation: 2 C H O + 65O14 14CO +14H2O 2 2 ? mol CO = 22.72 mol C H O 6 (14mo4 CO / 2 mol C 2 O )] = 16.6 m14 C4 2 Chemical Reactions in Solutions Most chemical reactions take place in solution partly because mixing the reactants helps to achieve the close contact between atoms, ions, or molecules necessary for the reaction to occur. Solvent determines whether the solution exists as a solid, liquid, or a gas. However, for learning purposes, the aqueous (aq) solution will be used as a solvent. Solute is the material being dissolved by the solvent. Aqueous reactions can be grouped into three general categories: Precipitation reactions, acid-base neutralization reactions and oxidation-reduction (red ox) reactions. Molarity Molarity is a solution property defined as the number of moles solute per liter of solution Molarity (M) = (amount of solute, in moles / volume of solution, in liters) The term M stands for the term molar or mol/L. Problem: A solution is prepared by dissolving 25.0 mL ethanol, C H OH (d= 02785 g/mL), in enough water to produce 250.0mL solution. What is the molarity of ethanol in the solution? Solution: To determine the molarity, we must first find how many moles of ethanol exists in a 25.0 mL sample. ? mol C H OH = [25.0 mL C H OH x ( 0.789g C H OH/ 1 mL C H OH) x ( 1 mol C H OH/ 46.07 g C H OH)] 2 5 2 5 2 5 2 5 2 5 2 5 = 0.428 mol C H 2H5 Now apply the definition of molarity to this answer. Thus divide the moles of C H OH by the 2ol5tion size, 0.2500L. Molarity = (0.428 mol C H OH2/50.2500 L soln) = 1.71M C H OH 2 5 Preparation of a Solution 1. Weigh the solid sample. 2. Dissolve it in a volumetric flask partially filled with solvent. 3. Carefully fill to the mark. Solution Dilution M xiV = i x V f f Problem: A particular analytical chemistry procedure requires 0.0100 M K CrO . What volume 2f 0.450 M K C2O mus4 be diluted with water to prepare 0.250L of 0.0100 M K CrO ? 2 4 Solution: ? mol K C2O = [4.250 L soln x (0.0100 mol K CrO /1L so2n)] =40.00250 mol K CrO 2 4 ? L soln = [0.00250 L soln x ( 1L soln / 0.250 K CrO ] 2 0.0140 L soln OR V = V x M/M = [250.0mL x (0.0100 M/ 0.250 M)] = 10.0 mL i f f i Determining Limiting Reactant The reactant that is completely consumed is called the limiting reactant. The reactant that remains after a reaction is done is said to be in excess. Problem: Phosphorus trichloride, PCl , is a c3mmercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products.
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