lect_9oct27f09

lect_9oct27f09 - Outline 9 27-OCT-2009 1 Review Questions...

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1 Outline 9 27-OCT-2009 1. Review – Questions? % yield (do stoichiometry to find what could get then compare with what actually get) # mol = V*M 2. Solution preparation 3. Solution Stoichiometry # mol = V*M , another way to get # mol Next exam will have solution stoichometry problem(s)! mass→ mol & volume → mol, ICE
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2 Name______ Sample Quiz 11 TA Name______ 1. (2) Calculate the number of mol of NaOH in 25.0 mL of 0.0400 M NaOH. 2. (3) Calculate mass of NaOH (40.0 g/mol) in 25.0 mL of 0.0400 M NaOH.
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3 REVIEW: Solutions, Dilution Molarity and volume are inversely proportional. So, adding water makes the solution less concentrated. M i *V i = M f *V f Adding water increases final volume, V f , and decreases final molarity, M f . initial moles solute = final moles solute
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4 REVIEW: Solutions, Dilution; Ex. 2(s1) Did Ex. 1 in lecture #8 (slide 36). Ex: p. 3.97(b) , p. 129. What volume of 1.03 M calcium chloride soln must be diluted with water to make 750.0 mL of 2.66*10 -2 M chloride solution? Diagram. Given: 750.0 mL of 2.66*10 -2 M Cl Find: Volume 1.03 M CaCl 2 needed Soln: Use # mol Cl - before = # mol Cl - after or M i V i = M f V f or (V i M i = V f M f ) What volume of water is needed? (Enough)
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5 REVIEW: Solutions, Dilution; Ex. 2 (s2) Ex : p. 3.97(b) , p. 129. Given: 750.0 mL of 2.66*10 -2 M Cl - Find: Volume 1.03 M CaCl 2 needed Soln: Use # mol Cl - before = # mol Cl - after or V i *M i = V f * M f & (2 mol Cl - / 1 mol CaCl 2 ) V * (1.03 mol CaCl 2 /L)* (2 mol Cl - /mol CaCl 2 ) = 0.750 L * (2.66*10 -2 mol Cl - / L) Rearrange V = 0.750* [2.66*10 -2 / (2* 1.03)]* L = 9.68 mL of 1.03 M CaCl 2 soln
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6 REVIEW: Solutions, Dilution; Ex. 2 (s3) Ex (cont. 2): p. 3.97(b) , p. 129. Given: 750.0 mL of 2.66*10 -2 M Cl - Find: Volume 1.03 M CaCl 2 needed Soln: Use # mol Cl - before = # mol Cl - after or M i V i = M f V f So, measure out 9.68 mL (buret) of 1.03 M CaCl 2 soln and add ENOUGH water to make 750.0 mL soln. Volume Solvent ≠ Volume Solution (No law of conservation of volume!)
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7 Solution Stoichiometry; Ex. 1 (vol A→ # g B) Prob. How many mL of 0.383 M HCl are needed to react with 16.2 g CaCO 3 (100.1 g/mol)? Given: 0.383 M HCl and 16.2 g CaCO 3 . Find: # mL of 0.383 M HCl needed Soln: (a) need balanced equation 2 HCl(aq)+CaCO 3 (s) →CaCl 2 (aq)+CO 2 (g)+H 2 O(l) Heart of problem is (2mol HCl/mol CaCO 3 ) (b) Use mols. (# g A → # mol A →#mol B → # mL B soln) # mL = 16.2 g CaCO 3 * (mol CaCO 3 /100.1 g CaCO 3 )* (2mol HCl/mol CaCO 3 )* (1 L/0.383 mol HCl)* (1000 mL/L) = ? mL soln.
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8 Solution Stoichiometry, Ex. 2 (vol A→ # g B) Prob: How many grams NaH 2 PO 4 (120. g/mol) are needed to react with 38.74 mL of 0.275 M NaOH? Given:
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lect_9oct27f09 - Outline 9 27-OCT-2009 1 Review Questions...

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