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Chapter 9. Stoichiometry The calculation of quantities in chemical reactions The calculation of quantities in chemical reactions.

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Presentation on theme: "Chapter 9. Stoichiometry The calculation of quantities in chemical reactions The calculation of quantities in chemical reactions."— Presentation transcript:

1 Chapter 9

2 Stoichiometry The calculation of quantities in chemical reactions The calculation of quantities in chemical reactions

3 Particles N 2(g) + 3H 2(g) 2NH 3(g) N 2(g) + 3H 2(g) 2NH 3(g) 1 molecule of N 2 reacts with 3 molecules of H 2 to produce 2 molecules of NH 3 1 molecule of N 2 reacts with 3 molecules of H 2 to produce 2 molecules of NH 3 Ratio : 1:3:2 particles Ratio : 1:3:2 particles

4 Moles 1 mole = 6.02 x representative particles Based on the reaction – mole ratio is 1:3:2 * Coefficients are the relative # of moles

5 Mass Law of conservation of mass Law of conservation of mass 1:3 2 28g + 6 g 34 g N = 2x14 H = 2x3 NH 3 = 17x2

6 Volume Assume STP Assume STP 22.4 L/mol 22.4 L/mol

7 Example 2H 2 S + 3O 2 2SO 2 + 2H 2 O

8 Mass of Reactants 2 mol x 34.1 g = 68.2g mol mol 3 mol x 32 g = 96g mol mol g g

9 Mass of Products 2 mol x 64.1g = g mol mol 2 mol x 18g = 36 g mol mol g g

10 Volume of gases at STP Reactants Reactants 2 mol H 2 S x 22.4 L = 44.8 L H 2 S 1 mol 1 mol 3 mol O 2 x 22.4 L = 67.2 L O 2 1 mol 1 mol

11 Products Products 2 mol SO 2 x 22.4 L = 44.8 L SO 2 1 mol 1 mol 2 mol H 2 O x 22.4 L = 44.8 L H 2 O 1 mol 1 mol

12 Mole to Mole Calculations How many moles of ammonia are produced when 0.06 mol of nitrogen reacts with hydrogen? How many moles of ammonia are produced when 0.06 mol of nitrogen reacts with hydrogen?

13 N 2 + 3H 2 2NH mol N 2 x 2mol NH mol N 2 x 2mol NH 3 1 mol N 2 1 mol N 2 =.12mol NH 3 =.12mol NH 3

14 Example Aluminum Oxide is formed from aluminum and oxygen a) Write a balanced equation a) Write a balanced equation 4Al + 3O 2 2Al 2 O 3 4Al + 3O 2 2Al 2 O 3

15 2.3 mol Al 2 O 3 x 4 mol Al 2.3 mol Al 2 O 3 x 4 mol Al 2 mol Al 2 O 3 2 mol Al 2 O 3 = 4.6mol Al = 4.6mol Al b) How many moles of Al are needed to form 2.3 mol of Al 2 O 3 ?

16 .84mol Al x 3mol O 2 4mol Al 4mol Al c) How many moles of Oxygen are required to react completely with 0.84 mol of Al ? =.63 mol O 2

17 17.2 mol O 2 x 2mol Al 2 O 3 3mol O 2 3mol O 2 = mol Al 2 O 3 = mol Al 2 O 3 d) Calculate the # of moles of Al 2 O 3 formed when 17.2 mol of O 2 reacts with Al ?

18 Mass – Mass Calculations - GFM - Ex) H 2 1 mol or 2g 1 mol or 2g 2g 1 mol 2g 1 mol

19 Example Calculate the number of grams of ammonia produced by the reaction of 5.40 g of hydrogen with nitrogen. Calculate the number of grams of ammonia produced by the reaction of 5.40 g of hydrogen with nitrogen. N 2(g) + 3H 2(g) 2NH 3(g) N 2(g) + 3H 2(g) 2NH 3(g)

20 5.40g H 2 x 1mol H 2 x 2mol NH 3 x 17g NH 3 2g H 2 3mol H 2 1mol NH 3 2g H 2 3mol H 2 1mol NH 3 = 30.6g NH 3 = 30.6g NH 3 These problems are solved the same way as mole – mole problems

21 Steps 1) Change mass to moles G (given) 2) Change moles of G to moles of W 3) Change moles of W to grams of W

22 Example 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O a) How many grams of O 2 are required to burn 13g of C 2 H 2 ? = 40.0g O 2

23 13g C 2 H 2 x 1mol C 2 H 2 x 4mol CO 2 x 44g CO 2 13g C 2 H 2 x 1mol C 2 H 2 x 4mol CO 2 x 44g CO 2 26g C 2 H 2 2mol C 2 H 2 1mol CO 2 26g C 2 H 2 2mol C 2 H 2 1mol CO 2 = 44g CO 2 = 44g CO 2 b) How many grams of CO 2 and H 2 O are produced when 13g of C 2 H 2 react with the oxygen.

24 Continued 13g C 2 H 2 x 1mol C 2 H 2 x 2mol H 2 O x 18g H 2 O 26g C 2 H 2 2mol C 2 H 2 1mol H 2 O 26g C 2 H 2 2mol C 2 H 2 1mol H 2 O = 9g H 2 O = 9g H 2 O

25 Other Stoichiometric Calculations 1) Mass – Volume 2) Volume – Volume 3) Particle – Mass

26 Example Tin plus hydroflouric acid yields Tin(II)flouride and hydrogen gas Tin plus hydroflouric acid yields Tin(II)flouride and hydrogen gas Sn (s) Sn (s) + 2HF (g) 2HF (g) SnF 2(s) SnF 2(s) + H 2(g)

27 966g SnF 2 a) How many grams of SnF 2 can be made by reacting 7.42 x molecules of HF with tin?

28 4.42 L H 2 b) How many L of H 2 (STP) are produced by reacting 23.4g of Sn with HF?

29 28.4 L HF c) How many L of HF are needed to produce 14.2 L of H 2 (STP) ?

30 1.08 x molecules H 2 d) How many molecules of H 2 are produced by the reaction of Sn with 80 L of HF (STP) ?

31 Limiting Reagent Example Example 1lb of swiss & 1lb of pastrami and 14 slices of bread 1lb of swiss & 1lb of pastrami and 14 slices of bread Limiting reagent - bread Limiting reagent - bread Excess reagent – luncheon meatsExcess reagent – luncheon meats You can only make 7 sandwichesYou can only make 7 sandwiches

32 Example Sodium Chloride is prepared by the reaction of sodium metal with chlorine gas Sodium Chloride is prepared by the reaction of sodium metal with chlorine gas 2Na (s) + Cl 2(g) 2NaCl (s) 2Na (s) + Cl 2(g) 2NaCl (s) What will occur when 6.70mol of Na reacts with 3.20 mol of Cl 2 ? What will occur when 6.70mol of Na reacts with 3.20 mol of Cl 2 ?

33 a) What is the limiting reagent ? Cl 2 is the limiting reagent Cl 2 is the limiting reagent. = 3.35 mol Cl 2 required * 3.35 mol Cl 2 are required to completely react with 6.70mol Na but you only have 3.20 mol Cl mol Na x 1mol Cl 2 2mol Na 2mol Na

34 = 6.40 mol NaCl b) How many moles of NaCl are produced? 3.20mol Cl 2 Cl 2 x 2mol NaCl 1mol Cl 2

35 .30mol Na is excess = 6.40mol Na c) How much of the excess reagent remains un-reacted ? 3.20mol Cl 2 Cl 2 x 2mol Na 1mol Cl mol Na 6.70mol Na mol Na

36 Example C 2 H 4 + 3O 2 2CO 2 + 2H 2 O C 2 H 4 + 3O 2 2CO 2 + 2H 2 O If 2.70 mol C 2 H 4 reacts with 6.30 mol 0 2 If 2.70 mol C 2 H 4 reacts with 6.30 mol 0 2 a) What is the limiting reagent?

37 the limiting reagent is O 2 2.7mol C 2 H 4 C 2 H 4 x 3mol O 2 O 2 = 8.1mol O2O2O2O2 1mol C2H4C2H4C2H4C2H4 * You need 8.1mol O 2 to react with 2.7 mol C 2 H 4

38 = 4.2mol H 2 O b) Calculate moles of H 2 O produced 6.3 mol O 2 O 2 x 2mol H2OH2OH2OH2O 3mol O2O2O2O2

39 = 2.1mol C 2 H 4 c) Calculate moles of excess reagent 6.30mol O 2 O 2 x 1mol C2H4C2H4C2H4C2H4 3mol O2O2O2O2

40 0.60mol C 2 H 4 Final Step Final Step 2.7mol C2H4C2H4C2H4C2H mol C2H4C2H4C2H4C2H4

41 Example a) How many grams of H 2 can be produced when 4g HCl is added to 3g of Mg?

42 =0.11g H2H2H2H2 4g HCl x 1mol HCl = 0.11mol HCl 4g HCl x 1mol HCl = 0.11mol HCl 36g HCl 36g HCl 3g Mg x 1mol Mg =.12mol Mg 24.3g Mg.11mol HCl x 1mol H 2 H 2 x 2g H2H2H2H2 2mol HCl 1mol H2H2H2H2

43 = 1.23 L H 2 b) Assuming STP, what is the volume of H 2 ? 0.11g H 2 H 2 x 1mol x 22.4L 2g 1mol

44 Percent Yield Theoretical Yield – the maximum amount of product that could be formed from a given amount of reactant Theoretical Yield – the maximum amount of product that could be formed from a given amount of reactant

45 Actual Yield – Amount of product that forms when the reaction is carried out in the lab. Actual Yield – Amount of product that forms when the reaction is carried out in the lab.

46 Percent Yield – Ratio of actual yield to theoretical yield Percent Yield – Ratio of actual yield to theoretical yield % Yield = actual yield x 100 % Yield = actual yield x 100 theoretical yield theoretical yield

47 Example CaCO 3 CaO + CO 2 CaCO 3 CaO + CO 2 *Calculate the % yield of CaO if 24.89g CaCO 3 is heated to give 13.1g of CaO

48 = 13.9g CaO 24.8g CaCO 3 x 1mol CaCO 3 x 24.8g CaCO 3 x 1mol CaCO 3 x 100.1g CaCO g CaCO 3 1mol CaO x 56.1g CaO 1mol CaO x 56.1g CaO 1mol CaCO 3 1mol CaO 1mol CaCO 3 1mol CaO

49 94.2% Percent Yield x 100 =

50 Example What is the % yield of copper if 3.74g of copper is produced when 1.87g of Aluminum is reacted with an excess of copper(II)sulfate? What is the % yield of copper if 3.74g of copper is produced when 1.87g of Aluminum is reacted with an excess of copper(II)sulfate?

51 2Al + 3CuSO 4 Al 2 (SO 4 ) 3 + 3Cu *3.74g Cu *3.74g Cu 1.87g Al x 1mol Al x 3mol Cu x 63.54g 26.98g Al 2mol Al 1mol Cu 26.98g Al 2mol Al 1mol Cu = 6.61g Cu = 6.61g Cu

52 Percent Yield 3.74g Cu x 100 = 3.74g Cu x 100 = 6.61g Cu 6.61g Cu 56.6 %


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