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) + C 6 H 8 O 8 (aq) + 2Br 2 (aq)  4HBr(aq) + C 6 H 6 O 6 (aq) + 2H 2 O The vitamin C in a 1.00 g chewable tablet requires 28.28 mL of 0.102 M Br 2 solution for titration to the equivalence point. How many grams of vitamin C (208 g/mol) are contained in the tablet? mols Br 2 mols C 6 H 8 O 8 g C 6 H 8 O 8 0.102 mols Br 2  L x 0.02828 L  1 x 1 mols C 6 H 8 O 8  2 mol Br 2 x 208 g C 6 H 8 O 8  mol C 6 H 8 O 8 = 0.300 g C 6 H 8 O 8

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) + C 6 H 8 O 8 (aq) + 2Br 2 (aq)  4HBr(aq) + C 6 H 6 O 6 (aq) + 2H 2 O The vitamin C in a 1.00 g chewable tablet requires 28.28 mL of 0.102 M Br 2 solution for titration to the equivalence point. How many grams of vitamin C (208 g/mol) are contained in the tablet? mols Br 2 mols C 6 H 8 O 8 g C 6 H 8 O 8 0.102 mols Br 2  L x 0.02828 L  1 x 1 mols C 6 H 8 O 8  2 mol Br 2 x 208 g C 6 H 8 O 8  mol C 6 H 8 O 8 = 0.300 g C 6 H 8 O 8

4. A 50.00 mL sample of solution containing Fe 2+ is titrated with 0.0216 M KMnO 4 solution. The solution required 20.65 mL of KMnO 4 solution to oxidize all of the Fe 2+ ions to Fe 3+ by the reaction: 8H + (aq) + MnO 4 1- (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) +5Fe 3+ (aq) + 4H 2 O 1 mols KMnO 4 mols MnO 4 1- m ols Fe 2 + M Fe 2+ 0.0216 mols KMnO 4  L x 0.02065 L  1 x 1 mols MnO 4 1-  1 mol KMnO 4 x 5 mols Fe 2+  1 mol MnO 4 1- 1  0.0500 L 0.0446 M Fe 2+ x= 2

5. A 50.00 mL sample of solution containing Fe 2+ is titrated with 0.0216 M KMnO 4 solution. The solution required 20.65 mL of KMnO 4 solution to oxidize all of the Fe 2+ ions to Fe 3+ by the reaction: 8H + (aq) + MnO 4 1- (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) +5Fe 3+ (aq) + 4H 2 O 1 mols KMnO 4 mols MnO 4 1- m ols Fe 2 + M Fe 2+ 0.0216 mols KMnO 4  L x 0.02065 L  1 x 1 mols MnO 4 1-  1 mol KMnO 4 x 5 mols Fe 2+  1 mol MnO 4 1- 1  0.0500 L 0.0446 M Fe 2+ x= 2

6. The Limiting Reagent 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. ®is totally consumed during the chemical reaction. ­determines,i.e. limits, the quantity of product(s) that will be obtained. ¬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.

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

8. CaC 2 (s) + 2H 2 O Ca(OH) 2 (aq) +C 2 H 2 (g) How many grams of C 2 H 2 (g) will be formed from the reaction of 24.0 g CaC 2 (s) and 18.0 g H 2 O ?

9. Empirical Formula The simplest whole- numbered (3) ratio (2) of numbers of mols of atoms (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 o C 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 C x H y was vaporized and, at 0.00 o C 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. C(s) + 2 H 2 O(l) CO 2 + 2 H 2 (g) How many grams of hydrogen gas are produced when 18 g C reacts with 27 g H 2 O?

13. From the balanced chemical equation: 1 mol C 0.50 mols C  or  2 mols H 2 O mol H 2 O C(s) + 2H 2 O  CO 2 (aq) +2H 2 (g) Conclusion: H 2 O is in the limiting reagent, C is in excess From Provided Reaction stoichiometry 1 mol C 0.50 mols C  vs  1 mols H 2 O 1 mol H 2 O Compare: mol H 2 O 27g H 2 O x  = 1.5 mol H 2 O 18.0 g H 2 O mol C 18.0g C x  = 1.5 mol C 12.0 g C Provided:

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

15. Al 2 (SO 4 ) 3 (aq) How many grams of precipitate will be formed when 308.6 mL of 0.324 M Al 2 (SO 4 ) 3 is poured into 432. mL of 1.157 M NaOH? +NaOH Al(OH) 3 Na 2 SO 4 (aq)+(s)23 6

16. How many grams of N 2 O(g) will be produced when 14.0 g N 2 (g) reacts with 30.0 g H 2 O(g)? N 2 (g) + H 2 O(g) N 2 O(g) + NH 3 (l) How many grams H 2 (g) will be formed when 2.16 g Al react with 2.92 g HCl (in aqueous solution)? 2 Al(s) + 6HCl(aq)  2 AlCl 3 (aq) + 3H 2 (g)

17. Al 4 C 3 (s) + H 2 O Al(OH) 3 (s) + CH 4 (g) How many grams of CH 4 (g) will be formed from the reaction of 14.4 g Al 4 C 3 (s) and 18.0 g H 2 O ?

18. Which is the limiting reagent? From the balanced chemical reaction: 1 mol Al 4 C 3 (s) 0.0833 mol Al 4 C 3 (s)  =  12 mols H 2 O 1 mols H 2 O Provided: mol Al 4 C 3 14.4 g All 4 C 3 (s) x  = 0.100 mol Al 4 C 3 144 g Al 4 C 3 mol H 2 O 18g H 2 O x  = 1.0 mol H 2 O 18.0 g H 2 O Compare: From Provided Reaction stoichiometry 0.100 mol Al 4 C 3 0.0833 mol Al 4 C 3  vs  1.0 mol H 2 O Al 4 C 3 (s) is in excess,, H 2 O is the limiting reagent Al 4 C 3 (s) + H 2 O  Al(OH) 3 (s) + CH 4 ( g)

19. Quantity of the product that will be obtained will be determined by the quantity of the limiting reagent provided. 1 mol H 2 O (l) 3 mol CH 4 16.0 g CH 4 18.0 g H 2 O (l) x  x  x  18.0 g H 2 O (l) 12 mols H 2 O (l) mol CH 4 = 4.0 g CH 4

20. Which is the limiting reagent? From the balanced chemical reaction: 1 mol Al 4 C 3 (s) 12 mols H 2 O Now, how many molsAl 4 C 3 (s) are needed to react with the number of mols of H 2 O provided? 1 mol H 2 O(s) 1 molsAl 4 C 3 (s) 18.0 g H 2 O(s) x x = 18 g H 2 O(s) 12 mols H 2 O (l) = 0.083 molsAl 4 C 3 (s) (needed) 18.0g H 2 O requires 0.083 mols Al 4 C 3 (s) for complete reaction. 14. 4 g Al 4 C 3 (s) is provided: 1 mol Al 4 C 3 (s)O 14.4 g Al 4 C 3 (s) x = 0.10 mols Al 4 C 3 (s) 144 g Al 4 C 3 (s) 0.10 mols Al 4 C 3 (s) (provided) > 0.083 mols Al 4 C 3 (s) H 2 O (needed ) Conclusion: Al 4 C 3 (s) is in excess, H 2 O 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 H 2 O (l) 3 mol CH 4 16.0 g CH 4 18.0 g H 2 O (l) x —————— x ————— x ————— 18.0 g H 2 O (l) 12 mols H 2 O (l) mol CH 4 = 4.0 g CH 4