Chemistry 11. Unit 7 - Stoichiometry

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1 1 Chemistry 11 Unit 7 - Stoichiometry

2 2 1. Coefficients of chemical equations In chapter 6, we have learned how to balance a chemical reaction by considering the laws of conservation of atoms and charges. The coefficients of a balanced chemical reaction thus tell us the relationship between the amounts of reactants used in the reaction and the amount of products produced by the reaction. Consider the following reaction: 2H 2 + O 2 2H 2 O It shows that two hydrogen molecules and one oxygen molecule react to generate two molecules of water.

3 3 The coefficients of a chemical reaction in fact indicates the ratio of the reactants and products involved in the process. As long as the ratio is maintained, the reaction is balanced. That means for the previous example, it can be rewritten as 200 H O H 2 O H O H 2 O or 2 mol H mol O 2 (2 mol) H 2 O

4 4 Example: Consider the reaction equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) How many molecules of N 2 are required to react with 15 molecules of H 2? According to the balanced equation, 1 molecule of N 2 reacts with 3 molecules of H 2. Hence, the number of N2 molecules required is Number of N 2 molecules = 15 H 2 molecules 1 N 2 molecule 3 H 2 molecules = 5

5 5 Example: Consider the reaction equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) How many moles of NH 3 are produced when 18 mol of H 2 are reacted? According to the balanced equation, 2 moles of NH 3 are produced when 3 moles of H 2 react. Hence, when 18 moles of H 2 are reacted, the number of moles of NH 3 produced will be: # mol NH 3 produced = 18 mol H 2 2 mol NH 3 3 mol H 2 = 12 mol

6 6 2. Stoichiometry involving mass, moles, volume and molecules Stoichiometric calculations make use of the relationship between the amounts of reactants and products involved in a chemical reaction. Based on how much of chemical #1 is involved in the process, the amount of chemical #2 involved can be deduced. Although the stoichiometric ratio of a balanced chemical equation is given in terms of mole ratio, since mass, volume and number of molecules are related to mole (Chapter 6), stoichiometric calculations can be performed technically in any of these quantities if proper unit conversions are done.

7 7 Mole is the unit that should be used in all stoichiometric calculations. Therefore, if the given quantity is not in terms of moles, then a unit conversion must be performed first. Depending on the unit with which the final answer should be provided, a second conversion may be needed. In summary:

8 8 Example: Consider the balanced equation: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) What mass of C 3 H 8 is required to produce g of H 2 O? Recall the relationship: Hence, mass H 2 O mol H 2 O mol C 3 H 8 mass C 3 H 8 Mass C 3 H 8 = g 1 mol H 2O 18 g = 61.1 g 1 mol C 3H 8 4 mol H 2 O 44 g 1 mol C 3 H 8

9 9 Example: Consider the balanced equation: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) If a sample of propane is burned what mass of H 2 O is produced if the reaction also produces 50.0 L of CO 2 at STP? Recall the relationship: Hence, vol CO 2 mol CO 2 mol H 2 O mass H 2 O mass H 2 O = 50.0 L 1 mol CO L 4 mol H 2O 3 mol CO g 1 mol H 2 O = 53.6 g

10 10 Example: Consider the balanced equation: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) A sample of porous, gas-bearing rock is crushed and g of C 3 H 8 is extracted from the powdered rock. How many molecules of CO 2 are produced if the gas sample is burned in the presence of an excess of O 2? Recall the relationship: Hence, mass C 3 H 8 mol C 3 H 8 mol CO 2 mass CO 2 Molecules CO 2 = g 1 mol C 3H 8 44 g = molecules 3 mol CO 2 1 mol C 3 H molecules 1 mol CO 2

11 11 Practice: Consider the reaction equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) (a) What mass of NH 3 is formed from the reaction of 25.0 g of H 2? (b) What volume of N 2 at STP is required to react with 30.0 g of H 2? (c) How many molecules of NH 3 are produced from the reaction of 15.0 g of N 2?

12 12 3. Stoichiometry involving concentration Recall from chapter 6 that the concentration of a solution can be calculated by M = number of moles volume (inl) = n V Rearranging this expression yields: n = MV Hence, if either concentration or volume of a solution involved in a stoichiometric calculation is given, the corresponding number of moles can be computed. Note: DO NOT use 22.4 L as it is the molar volume of gas at STP only!

13 13 Example: A tablet of Tums has a mass of g of CaCO 3. What volume of stomach acid having [HCl] = M is neutralized by one piece of Tums? Recall the reaction between CaCO 3 and HCl: CaCO 3 + 2HCl CaCl 2 + CO 2 + H 2 O The amount of CaCO 3 in one tablet of Tums: mol CaCO 3 = g 1 mol = mol g The amount of HCl to be neutralized: 2 mol HCl mol HCl = mol CaCO 3 = mol 1 mol CaCO 3 The volume of stomach acid to be neutralized: mol vol HCl = M = 15 L

14 14 Example: What volume of CO 2 at STP is produced if 1.25 L of M HCl reacts with an excess of CaCO 3? The chemical equation is CaCO 3 + 2HCl CaCl 2 + CO 2 + H 2 O The amount of HCl reacted: mol HCl = M 1.25 L = mol By mole ratio, the number of moles of CO 2 produced: mol CO 2 = mol 1 mol CO 2 = mol 2 mol HCl Hence, the volume of CO 2 produced: vol CO 2 = mol 22.4 L = L

15 15 4. Excess and limiting reagents All the stoichiometric calculations performed in the previous sections assumed that species react completely (in other words, following the stoichiometry perfectly). This, however, is mostly not the case in reality because of the following reasons: (1) To make sure the second reagent (too expensive to waste) is reacted completely. (2) To be unavoidable since only a limited amount of the other reagent is available.

16 16 In these calculations, the species that exists in a lesser amount determines the amount of product that can be formed. This species is called the limiting reagent. Note that the limiting reagent is used up completely in a chemical reaction. Any reactants that remain at the end are called excess reagents. Species Limiting reagent Excess reagent Characteristics Used up in a reaction Determines the amount of product to be formed Not used up in a reaction Does not limit the amount of product to be formed

17 17 Example: If 20.0 g of H 2 reacts with g of O 2 according to the reaction: 2H 2 (g) + O 2 (g) 2H 2 O(g) Which reactant is in excess and by how much? To determine which of the reactants is in excess, assume one reactant as limiting and calculate the necessary amount of the other reactant in order to complete the reaction. (i) Assume H 2 as limiting: Mass of O 2 = 20.0 g 1 mol H 2 2 g 1 mol O 2 2 mol H g 1 mol O 2 = 160 g

18 18 (ii) Assume O 2 as limiting: Mass of H 2 = g 1 mol O g 2 mol H 2 1 mol O g 1 mol H 2 = g As seen, consuming all 20.0 g of H 2 requires 160 g of O 2, but there is only g available. Meanwhile, only 12.5 g of H 2 is needed to react completely with g O 2, yet there is 20.0 g of H 2 available. Hence, it can be concluded that O 2 is the limiting reagent, and H 2 is in excess. The amount of H 2 in excess: H 2 in excess = 20.0 g 12.5 g = 7. 5 g

19 19 Example: If 56.8 g of FeCl 2, 14.0 g of KNO 3, and 40.0 g of HCl are mixed and allowed to react according to the reaction: 3FeCl 2 + KNO 3 + 4HCl 3FeCl 3 + NO + 2H 2 O + KCl (a) which reactant is the limiting reactant? (b) how many grams of each excess reagent remain? If more than two species are involved in a reaction, the previous approach will become less convenient. Indeed, there is a second method of finding the limiting reagent, in which the expected amounts of products assuming each reagent being limiting are calculated and compared. The one that yields the least amount will be the limiting reagent.

20 20 (a) Let find the mass of NO, (i) assuming FeCl 2 is limiting: 56.8 g FeCl 2 1 mol FeCl g 1 mol NO 30.0 g 3 mol FeCl 2 1 mol NO = 4.48 g (ii) assuming KNO 3 is limiting: 14.0 g KNO 3 1 mol KNO g (iii) assuming HCl is limiting: 40.0 g HCl 1 mol HCl 36.5 g Hence, KNO 3 is the limiting agent. 1 mol NO 30.0 g 1 mol KNO 3 1 mol NO = g 1 mol NO 4 mol HCl 30.0 g 1 mol NO = 8.22 g

21 21 (b) The amounts of excess reagents that remain can be found if how much of each is consumed is known. For FeCl 2 : 14.0 g KNO 3 1 mol KNO g 3 mol FeCl 2 1 mol KNO g 1 mol FeCl 2 = 52.7 g Mass FeCl 2 remains = 56.8 g 52.7 g = 4. 1 g For HCl: 14.0 g KNO 3 1 mol KNO g 4 mol HCl 36.5 g 1 mol KNO 3 1 mol HCl = 20.2 g Mass HCl remains = 40.0 g 20.2 g = g

22 22 Practice: Ammonia reacts with oxygen as shown in the following equation: 4NH 3 + 3O 2 2N 2 + 6H 2 O How many grams of H 2 O can be formed from a reaction mixture containing 35.0 g of NH 3 and 50.0 g of O 2?

23 23 5. Percentage yield and percentage purity Using stoichiometry, we can always calculate the amount of product which is to be obtained when all things go perfectly. It is called the expected yield. However, in reality it is not easy to achieve this level of performance. Mostly the amount of product actually obtained in an experiment (called actual yield) is smaller than the expected yield. The ratio of these quantities is called percentage yield: Percentage yield = Actual yield Expected yield 100%

24 24 There are two reasons that account for the reduced yield of an experiment: (1) The reactants are not pure; so not all reactants may react. (2) Some products may be lost during separation. The purity of a substance, called percentage purity, can be calculated as: Percentage purity = mass of pure reactant mass of impure reactant 100%

25 25 Example: When 15.0 g of CH 4 is reacted with an excess of Cl 2 according to the reaction: CH 4 + Cl 2 CH 3 Cl + HCl A total of 29.7 g of CH 3 Cl is formed. What is the percentage yield of the reaction? The expected yield is: 15.0 g CH 4 1 mol CH g 1 mol CH 3Cl 1 mol CH g 1 mol CH 3 Cl = g The percentage yield is: percetage yield = 29.7 g g 100% = 62.7%

26 26 Example: What mass of K 2 CO 3 is produced when 1.50 g of KO 2 is reacted with an excess of CO 2 according to the reaction 4KO 2 + 2CO 2 2K 2 CO 3 + 3O 2 if the reaction has a 76.0% yield? The expected yield is: 1.50 g 1 mol KO g The actual mass is: 2 mol K 2CO 3 4 mol KO g 1 mol K 2 CO 3 = g actual mass = g 0.76% = 1.11 g

27 27 Example: What mass of CuO is required to make 10.0 g of Cu according to the reaction 2NH 3 + 3CuO N 2 + 3Cu + 3H 2 O if the reaction has 58% yield? The expected yield is Mass of CuO needed is 10.0 g 1 58 % = g g 1 mol Cu 63.5 g 3 mol CuO 3 mol Cu 79.5 g 1 mol CuO = 21.6 g

28 28 Practice: When 25.0 g of H 2 are reacted with excess N 2 according to the reaction: N 2 + 3H 2 2NH 3 a total of g of NH 3 is formed. What is its percentage yield?

29 29 6. Titration The concentration of a solution can be determined by an experiment called titration. In titration, an unknown solution of precisely known volume (called analyte) is allowed to react with a chemical whose concentration is known accurately (called titrant or standard). An indicator is used to show when the reaction is completed (called the equivalence point). By stoichiometry, the number of moles of the unknown, and in turn its concentration, can be found.

30 30 A typical setup of a titration experiment:

31 31 Example: If 19.8 ml of H 3 PO 4 with an unknown molarity reacts with 25.0 ml of M KOH, what is the molarity of the H 3 PO 4 solution? H 3 PO 4 and KOH react according to the following equation: H 3 PO 4 + 3KOH K 3 PO 4 + 3H 2 O moles of KOH = L M = mol Hence, moles of H 3 PO 4 = mol 1 mol H 3PO 4 3 mol KOH = mol and the concentration of H 3 PO 4 is: mol H 3 PO 4 = L = M

32 32 Practice: A 10.0 ml sample of a saturated solution of Ca(OH) 2 is titrated with 23.5 ml of M HCl. What is the molarity of the saturated Ca(OH) 2 solution? How much Ca(OH) 2 should be dissolved in ml of water to make such a saturated solution?

Stoichiometry. A. The Meaning of Coefficients in a Reaction Equation 1. Consider the following reaction: 200 H O H 2 O or

Stoichiometry. A. The Meaning of Coefficients in a Reaction Equation 1. Consider the following reaction: 200 H O H 2 O or Stoichiometry A. The Meaning of Coefficients in a Reaction Equation 1. Consider the following reaction: 2H 2 + O 2 2H 2 O The coefficients in the equation tell us that two hydrogen molecules react with