1. Stoichiometry Pre-AP Chemistry Unit 7, Chapter 8

2. 8.1 Into to Stoichiometry Stoichiometry - the calculation of quantities in chemical reactions Finding amounts in chemistry is kind of like finding amounts for a recipe.

3. CO(g) + O 2 (g)  CO 2 (g) RP moles mass volume of gas STP atoms Interpreting chemical equations: 2 molecules 2 moles 56.02g 44.8L 4 atoms 1 molecule2 molecules 1 mole 2 moles 32.00g 22.4L 2 atoms 88.02g 44.8L 6 atoms 22 Balance First! Remember calculate mass to hundredths place! Mass and number of atoms are always conserved in every chemical reaction. Moles, molecules and volumes may or may not be conserved.

4. Moles of substance given Mass of substance given R.P. of substance given Volume of substance given Atoms, formula units, molecules grams L mol 1 mol gfm 1 mol 22.4 L 1 mol 6.02 x 10 23 R.P. of substance wanted Atoms, formula units, molecules Mass of substance wanted grams Volume of substance wanted L Moles of substance wanted mol 22.4 L 1mol gfm 1mol 6.02 x 10 23 1mol Coefficients from balanced equation! Always follow the road map… Do not make up new paths! mol wanted mol given

5. Mole-Mole Calculations Mole-Mole Calculations We can relate moles of reactants to moles of products, creating “mole ratios” using the coefficients from our balanced equation. These will be our conversion factors in our dimensional analysis.

6. How many moles of aluminum are needed to form 2.3 moles of aluminum oxide? 4.6 mol Al 4Al + 3O 2  2Al 2 O 3 Reaction: Given/want: x mol2.3mol 2.3 mol Al 2 O 3 4 mol Al 2 mol Al 2 O 3 =

7. How many moles of oxygen are required to react completely with 0.84 mol of Al? 0.63 mol O 2 4Al + 3O 2  2Al 2 O 3 Reaction: Given/want: 0.84 mol x mol 0.84 mol Al3 mol O 2 4 mol Al =

8. Calculate the # of moles of aluminum oxide formed when 17.2 mol of oxygen reacts with Al. 11.5 mol Al 2 O 3 4Al + 3O 2  2Al 2 O 3 Reaction: Give/want:17.2 mol x mol 2 mol Al 2 O 3 = 17.2 mol O 2 3 mol O 2

9. Measurement in Lab In lab, we measure grams, not moles, so we need to use gfm, (the mass of a compound from the periodic table) as a conversion factor.

10. Steps to a stoichiometry problem:(mass-mass) A. Write a balanced chemical equation. Identify knowns and unknowns. B. Convert the given mass to moles by dividing by the gfm. C. Convert moles of the first substance to moles of the second substance by using the mole ratio (coefficients from the balanced equation). Mole ratio = moles “wanted” moles “given” D. Convert moles of the second substance into grams by multiplying by the gfm.

11. Calculate the number of grams of solid sodium oxide that will be produced when 115 g of solid Na react with excess O 2. = 155g Na 2 O A.Write a balanced chemical equation. Identify knowns and unknowns 4Na + O 2  2Na 2 O Reaction: Give/want: 115g x g Means more than enough 115g Na1 mol Na2 mol Na 2 O61.98g Na 2 O 22.99g Na4 mol Na 1 mol Na 2 O C.Convert moles of the first substance to moles of the second substance by using the mole ratio (coefficients from the balanced equation). Mole ratio = moles “wanted” moles “given” B.Convert the given mass to moles by dividing by the gfm. D.Convert moles of the second substance into grams by multiplying by the gfm. BC D A

12. 5. Mass-mass calculations: a)How many grams of oxygen are required to “burn”13.0 g of C 2 H 2 ? = 39.9g O 2 2C 2 H 2 (g)+5O 2 (g)  4CO 2 (g) +2H 2 O(g) Reaction: Give/want: 13g x g 13.0g C 2 H 2 1 mol C 2 H 2 5 mol O 2 32.00g O 2 26.04 g C 2 H 2 2 mol C 2 H 2 1 mol O 2 BCD A

13. b)How many grams of CO 2 and grams of H 2 O are produced when 13.0 g of C 2 H 2 reacts with an excess amount of oxygen? = 43.9 g CO 2 2C 2 H 2 (g)+5O 2 (g)  4CO 2 (g) +2H 2 O(g) Reaction: Give/want: 13g x g 13.0g C 2 H 2 1 mol C 2 H 2 4 mol CO 2 44.01g CO 2 26.04 g C 2 H 2 2 mol C 2 H 2 1 mol CO 2 BCD A x g = 9.00g H 2 O 13.0g C 2 H 2 1 mol C 2 H 2 2 mol H 2 O18.02 g H 2 O 26.04 g C 2 H 2 2 mol C 2 H 2 1 mol H 2 O

14. Use answers from a) and b) to show that this equation obeys the law of conservation of mass. Mass of Products = Mass of Reactants Products: 43.9g CO 2 + 9.00g H 2 O 52.9g 2C 2 H 2 (g)+5O 2 (g)  4CO 2 (g) +2H 2 O(g) Reaction: Reactants: 13.0g C 2 H 2 + 39.9g O 2 52.9g Mass is CONSERVE D!

15. a) How many grams of SnF 2 can be made by reacting 7.42 x 10 24 molecules of HF with tin? = 966g SnF 2 Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) Reaction: Given/want: 7.42 x 10 24 molec x g 7.42 x 10 24 molec HF 1 mol HF1 mol SnF 2 156.71g SnF 2 6.02 x 10 23 molec HF 2 mol HF 1 mol SnF 2

16. b) How many liters of hydrogen (STP) are produced by reacting 23.4 g of Sn with HF? Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) Reaction: Given/want: 23.4 gx L = 4.42 L H 2 23.4g Sn1 mol Sn1 mol H 2 22.4L H 2 118.71g Sn1 mol Sn 1 mol H 2

17. c) How many liters of HF are needed to produce 14.2 L of hydrogen? (STP) Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) Reaction: Given/want: 14.2 Lx L = 28.4L HF 14.2L H 2 1 mol H 2 2 mol HF22.4 L HF 22.4 L H 2 1 mol H 2 1 mol HF

18. Volume to Volume Conversions In volume-volume problems, the coefficients in a balanced chemical equation tell you not only the relative number of moles, but also the relative volumes of interacting gases.

19. d) How many molecules of hydrogen are produced by the reaction of tin with 80.0 L of HF (STP)? Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) Reaction: Given/want: 80.0 Lx molec. = 1.08 x 10 24 molecules H 2 80.0L HF1 mol HF1 mol H 2 6.02 x 10 23 molec H 2 22.4L HF2 mol HF 1 mol H 2

20. LIMITING REACTANTS AND PERCENTAGE YIELD Chapter 8, section 3

21. Balanced chemical equations are like a chemist’s recipe! Limiting Reactant (or reagent) limits or determines the amount of product formed it is the reagent (or reactant) used up. Excess Reactant more than enough to react with the limiting reagent. is present in excess and will not be used up.

22. Let’s say you are making s’mores. You have 10 chocolate squares, 20 graham cracker squares, and 8 marshmallows. If each s’more requires 1 chocolate square, two graham cracker squares, and one marshmallow, what is the limiting reagent?

24. Suppose 3.8 mol hydrogen reacts with 2.2 mol of oxygen. A. What is the limiting reagent? H 2 is the limiting reagent 2H 2 + O 2  2H 2 O Reaction: Give/want: 3.8 mol2.2 mol = 3.8 mol H 2 O 3.8 mol H 2 2 mol H 2 O 2 mol H 2 = 4.4 mol H 2 O 2.2 mol O 2 2 mol H 2 O 1 mol O 2 Lower amount of H 2 O Limiting

25. B. How many moles of excess reagent are left over? 2.2 mol O 2 started - 1.9 mol O 2 used = 2H 2 + O 2  2H 2 O Reaction: Give/want: 3.8 molx mol used = 1.9 mol O 2 used 3.8 mol H 2 1 mol O 2 2 mol H 2 Since H 2 is limiting, O 2 is excess. We had 2.2 mol of oxygen, but only used 1.9 mol. 0.3 mol of O 2 left over

26. C. How many moles of water are produced? How many grams of water? (use the limiting reagent!) 2H 2 + O 2  2H 2 O Reaction: Give/want: 3.8 mol2.2 mol 3.8 mol H 2 O18.02g H 2 O 1 mol H 2 O = 68 g H 2 O 3.8 mol Using the moles of H 2 O determined in part A: Limiting

27. Ex. a) How many grams of hydrogen can be produced when 4.00 g of HCl is added to 3.00 g of Mg? 0.111g H 2 produced HCl is limiting. = 0.249 g H 2 Mg(s) +2HCl  MgCl 2 (aq) + H 2 (g) Reaction: Give/want: 3.00g x g 3.00g Mg1 mol Mg1 mol H 2 2.02 g H 2 24.31g Mg1 mol Mg 1 mol H 2 = 0.111g H 2 4.00g HCl1 mol HCl1 mol H 2 2.02g H 2 36.46g HCl2 mol HCl 1 mol H 2 4.00g Limiting

28. b) Assuming STP, what is this volume of H 2 ? = 1.22 L H 2 Mg(s) +2HCl  MgCl 2 (aq) + H 2 (g) Reaction: Give/want: 3.00g 0.110 g 0.110g H 2 1 mol H 2 22.4 L H 2 2.02g H 2 1 mol H 2 4.00g Limiting

29. Theoretical and Actual Yield Theoretical yield is obtained when an equation is used to calculate the amount of product that will form during a reaction. Actual Yield is the amount of product that forms when the reaction is carried out in the lab. always a calculated number. the maximum amount of product that could be formed from a given amount of reactant. an experimental value. often less than the theoretical yield.

30. Percent Yield Percent Yield is the ratio of the actual yield to the theoretical yield. % Yield = actual yield theoretical yield x 100

31. Ex. What is the % yield if 3.74 g of Cu is produced when 1.87 g Al is reacted with an excess of copper (II) sulfate? 2Al + 3CuSO 4  3Cu + Al 2 (SO 4 ) 3 Reaction: Give/want: 1.87 gx g = 6.61 g Cu 1.87g Al1 mol Al3 mol Cu63.55g Cu 26.98g Al2 mol Al 1 mol Cu % Yield = 3.74 g 6.61 g = 56.7% yield x 100

32. Ex. The percent yield for the reaction PCl 3 + Cl 2  PCl 5 is 83.2%. What mass of PCl 5 is expected from 73.7g of PCl 3 with excess chlorine? = 93.2 g (actual yield) PCl 3 + Cl 2  PCl 5 Reaction: Give/want: 73.7 gx g = 112 g PCl 5 (theoretical yield) 73.7g PCl 3 1 mol PCl 3 1 mol PCl 5 208.22g PCl 5 137.32g PCl 3 1 mol PCl 3 1 mol PCl 5 % Yield x g 112 g x 100 = 83.2 % yield = but only 83.2% yield