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Titrations

Titrations. Viatmin C is both an acid and a reducing agent. One method of determining the amount of Vit C in a sample is to carry out the following reaction via a titration: 2H + (aq) + C 6 H 8 O 8 (aq) + 2Br 2 (aq) 4HBr(aq) + C 6 H 6 O 6 (aq) + 2H 2 O

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Titrations

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  1. Titrations

  2. Viatmin C is both an acid and a reducing agent. One method of determining the amount of Vit C in a sample is to carry out the following reaction via a titration: 2H+(aq) + C6H8O8(aq) + 2Br2(aq) 4HBr(aq) + C6H6O6 (aq) + 2H2O The vitamin C in a 1.00 g chewable tablet requires 28.28 mL of 0.102 M Br2 solution for titration to the equivalence point. How many grams of vitamin C (208 g/mol) are contained in the tablet? mols C6H8O8 g C6H8O8 mols Br2 0.102 mols Br2  L 0.02828 L  1 1 mols C6H8O8 2 mol Br2 x x 208g C6H8O8 mol C6H8O8 x = 0.300 g C6H8O8

  3. Viatmin C is both an acid and a reducing agent. One method of determining the amount of Vit C in a sample is to carry out the following reaction via a titration: 2H+(aq) + C6H8O8(aq) + 2Br2(aq) 4HBr(aq) + C6H6O6 (aq) + 2H2O The vitamin C in a 1.00 g chewable tablet requires 28.28 mL of 0.102 M Br2 solution for titration to the equivalence point. How many grams of vitamin C (208 g/mol) are contained in the tablet? mols C6H8O8 g C6H8O8 mols Br2 0.102 mols Br2  L 0.02828 L  1 1 mols C6H8O8 2 mol Br2 x x 208g C6H8O8 mol C6H8O8 x = 0.300 g C6H8O8

  4. A 50.00 mL sample of solution containing Fe2+ is titrated with 0.0216 M KMnO4 solution. The solution required 20.65 mL of KMnO4 solution to oxidize all of the Fe2+ ions to Fe3+ by the reaction: 8H+(aq) + MnO41-(aq) + 5Fe2+(aq)  Mn2+(aq) +5Fe3+(aq) + 4H2O 1 mols MnO41- mols Fe2+ M Fe2+ mols KMnO4 2 0.0216 mols KMnO4 L 0.02065 L  1 1 mols MnO41- 1 mol KMnO4 x x 5 mols Fe2+  1 mol MnO41- 1 0.0500 L 0.0446 M Fe2+ x x =

  5. A 50.00 mL sample of solution containing Fe2+ is titrated with 0.0216 M KMnO4 solution. The solution required 20.65 mL of KMnO4 solution to oxidize all of the Fe2+ ions to Fe3+ by the reaction: 8H+(aq) + MnO41-(aq) + 5Fe2+(aq)  Mn2+(aq) +5Fe3+(aq) + 4H2O 1 mols MnO41- mols Fe2+ M Fe2+ mols KMnO4 2 0.0216 mols KMnO4 L 0.02065 L  1 1 mols MnO41- 1 mol KMnO4 x x 5 mols Fe2+  1 mol MnO41- 1 0.0500 L 0.0446 M Fe2+ x x =

  6. The Limiting Reagent • relative to the other reactant(s) of a chemical reaction, this reactant is present in less than the stoichiometrically equivalent amount i.e. you have less than you need to fully react with other reactants. • determines,i.e. limits, the quantity of product(s) that will be obtained. • is totally consumed during the chemical reaction. Reactants other than the limiting reagent are in excess (i.e. in excess of that amount required for stoi- chiometric equivalence with the limiting reagent). Some quantity of this (these) reactant(s) will remain after the reaction is complete.

  7. Example • Desire • Provided with: and Which, or , is the “limiting” supply?

  8. CaC2(s) + 2H2O Ca(OH)2(aq) +C2H2(g) How many grams of C2H2(g) will be formed from the reaction of 24.0 g CaC2(s) and 18.0 g H2O ?

  9. Empirical Formula The simplest whole-numbered (3) ratio (2) of numbersofmolsofatoms (1) in one mol of a compound.

  10. Mo(CO)x(s) Mo(s) + xCO(g) When a 2.200 g of Mo(CO)x is heated it decomposes producing 0.7809 g of Mo(s) and gaseous CO. The CO gas was found to occupy a volume of 1.2345 L at a temperature of 31.50 oC and a pressure of 751.1 mm Hg. What is the value of x? R= 0.0821 L-atm/mol K R = 62,400 mL-mm/mol K

  11. A compound with the general formula CxHy was vaporized and, at 0.00 oC and 760 mm Hg, was found to have a density of 5.0996 g per L. In a separate determination the elemental composition of the compound was found to be 84.118 % C and 15.882 % H. (1). Calculate the molar mass of the compound. (2). Calculate the empirical formula of the compound (3). What is the molecular formula of the compound? R= 0.0821 L-atm/mol K R = 62,400 mL-mm/mol K

  12. How many grams of hydrogen gas are produced when 18 g C reacts with 27 g H2O? C(s) + 2 H2O(l) CO2 + 2 H2(g)

  13. C(s) + 2H2O  CO2(aq) +2H2(g) From the balanced chemical equation: 1 mol C0.50 mols C or 2 mols H2O mol H2O Provided: mol C 18.0g Cx  = 1.5 mol C 12.0 g C mol H2O 27g H2O x  = 1.5 mol H2O 18.0 g H2O Compare: From ProvidedReaction stoichiometry 1 mol C0.50 mols C vs 1 mols H2O 1 mol H2O Conclusion: H2O is in the limiting reagent,C is in excess

  14. mol H2O(l) 2 mol H2 27.0 g H2 O(l) x  x  18.0 g H2O(l) 2 mols H2O(l) 2.02 g H2 x mol H2(g) 3.03 g H2 =

  15. How many grams of precipitate will be formed when 308.6 mL of 0.324 M Al2(SO4)3 is poured into 432. mL of 1.157 M NaOH? Al2(SO4)3(aq) + 6 NaOH 3 Na2SO4 (aq) + 2 Al(OH)3 (s)

  16. How many grams of N2O(g) will be produced when 14.0 g N2(g) reacts with 30.0 g H2O(g)? N2(g) + H2O(g) N2O(g) + NH3(l) How many grams H2(g) will be formed when 2.16 g Al react with 2.92 g HCl (in aqueous solution)? 2 Al(s) + 6HCl(aq)  2 AlCl3(aq) + 3H2(g)

  17. Al4C3(s) + H2O Al(OH)3(s) + CH4(g) How many grams of CH4(g) will be formed from the reaction of 14.4 g Al4C3(s) and 18.0 g H2O ?

  18. Al4C3(s) + H2O  Al(OH)3(s) + CH4(g) Which is the limiting reagent? From the balanced chemical reaction: 1 mol Al4C3(s) 0.0833 mol Al4C3(s) = 12 mols H2O 1 mols H2O Provided: mol Al4C3 14.4 g All4C3(s)x  = 0.100 mol Al4C3 144 g Al4C3 mol H2O 18g H2O x  = 1.0 mol H2O 18.0 g H2O Compare: From ProvidedReaction stoichiometry 0.100 mol Al4C30.0833 mol Al4C3 vs 1.0 mol H2O 1.0 mol H2O Al4C3(s) is in excess, , H2Ois the limiting reagent

  19. Quantity of the product that will be obtained will be determined by the quantity of the limiting reagent provided. 1 mol H2O(l) 3 mol CH4 16.0 g CH4 18.0 gH2O(l) x  x  x  18.0 g H2O (l) 12 mols H2O (l) mol CH4 = 4.0 g CH4

  20. Which is the limiting reagent? From the balanced chemical reaction: 1 mol Al4C3(s) 12 mols H2O Now, how many molsAl4C3(s) are needed to react with the number of mols of H2O provided? 1 mol H2O(s) 1 molsAl4C3(s) 18.0 gH2O(s)x x = 18 g H2O(s) 12 mols H2O (l) = 0.083 molsAl4C3(s) (needed) 18.0gH2Orequires 0.083 mols Al4C3(s) for complete reaction. 14. 4 g Al4C3(s) is provided: 1 mol Al4C3(s)O 14.4 g Al4C3(s) x = 0.10 mols Al4C3(s) 144 g Al4C3(s) 0.10 mols Al4C3(s) (provided) > 0.083 mols Al4C3(s) H2O (needed ) Conclusion: Al4C3(s) is in excess, H2O is the limiting reagent

  21. Quantity of the product that will be obtained will be determined by the quantity of the limiting reagent provided. 1 mol H2O(l) 3 mol CH4 16.0 g CH4 18.0 gH2O(l) x —————— x ————— x ————— 18.0 g H2O (l) 12 mols H2O (l) mol CH4 = 4.0 g CH4

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