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I. I.Stoichiometric Calculations Stoichiometry – Ch. 11.

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Presentation on theme: "I. I.Stoichiometric Calculations Stoichiometry – Ch. 11."— Presentation transcript:

1 I. I.Stoichiometric Calculations Stoichiometry – Ch. 11

2 Background on things you NEED to know how to do: 1. Name/write correct chemical formula 2. Write chemical equations 3. Balance chemical equations 4. Predict Products 5. Mole/mass conversions

3 StoichiometryStoichiometry o Stoichiometry uses ratios to determine relative amounts of reactants or products. o For example If you were to make a bicycle, you would need one frame and two tires. o 1 frame + 2 tires  1 bicycle o If I had 74 tires, what is the most # of bicycles I could make? 74 tires 2 tires 1 bicycle = 37 bicycles

4 Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

5 Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation can be used to determine expected amounts of products given amounts of reactants. 2 Mg + O 2  2 MgO

6 Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass -moles  grams Molarity - moles  liters soln Molar volume -moles  liters gas Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

7 Molar Volume at STP Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

8 Mole – Mole Conversions  The first type of problems we encounter will go between moles and moles. For this we need to use mole ratios.  Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide. Pb(NO 3 ) 2 + 2KI  2 KNO 3 + PbI 2 Mole ratio of potasium iodide to lead (II) iodide: 2 moles KI 1 mole PbI 2

9 Mole to Mole Problems b How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

10 Mole to Mass b We can also convert from moles to mass, and mass to moles b For Example: 4 Al + 3 O 2  2Al 2 O 3 b If you know how many grams of Al you start with, we can write a flow chart to show how to calculate the # of moles of oxygen need to fully react with the Al. b g Al  moles Al  moles of oxygen

11 Mass to Moles: 4 Al + 3 O 2  2Al 2 O 3 b If the reaction starts with.84 moles of aluminum, how many grams of aluminum oxide are produced?.84 mol Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.9 grams Al 2 O 3 = 42.8 grams Al 2 O 3 b 0.92 g of Aluminum oxide are produced from the reaction. How much aluminum was used up?.92 g Al 2 O 3 101.9 g Al 2 O 3 1 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al 1 mol Al 26.9 g Al =.49 grams Al

12 Mass to Mass b How many grams of silver will be formed from 12.0 g copper reacting with silver nitrate? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

13 Stoichiometry Problems – Mole-Mole b N 2 H 4 + N 2 O 4  N 2 + H 2 O b ? moles N 2 O 4 = 2.72 moles N 2 H 4 4 2 3 2.72 mol N 2 H 4 1 mol N 2 O 4 2 mol N 2 H 4 ? mol 2.72 mol = 1.36 mol N 2 O 4

14 Stoichiometry Problems – Mass/Mass b Sodium metal reacts with oxygen gas to produce solid sodium oxide. How many grams of sodium must react to produce 42.0 grams of sodium oxide? 4Na + O 2  2Na 2 O ? g 42.0 g 42.0 g Na 2 O 16.98 g Na 2 O 1 mol Na 2 O 2 mol Na 2 O 4 mol Na 1 mol Na 22.99 g Na = 22.99 g Na

15 Stoichiometry Problems – Mole/Mass b In photosynthesis, carbon dioxide and water react to form glucose, C 6 H 12 O 6 and oxygen gas. ___CO 2 + ___H 2 O  ___C 6 H 12 O 6 + ___O 2 If 15.6 grams of carbon dioxide react, how many moles of glucose will be produced? How many grams of carbon dioxide must react to produce 0.25 moles of glucose? 6 6 6 15.6 g CO 2 44.01 g CO 2 1 mol CO 2 6 mol CO 2 1 mol C 2 H 12 O 6 = 0.0591 mol C 2 H 12 O 6 44.01 g CO 2 6 mol CO 2 1 mol CO 2 0.25 mol C 2 H 12 O 6 = 66 g CO 2 1 mol C 2 H 12 O 6

16 Stoichiometry with Gases b If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction.

17 At STP 1 mol of a gas=22.4 L Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

18 Molar Volume b Hydrogen and chlorine gas react to produce hydrochloric acid. If 7.00 L of hydrogen gas react, how many liters of HCl gas are formed? H 2 (g) + Cl 2 (g)  2 HCl (g) 7.00 L H 2 1.0 L H 2 2.0 L HCl = 14.0 L HCl *only in all gas reactions!

19 Molar Volume b In the following reaction, if 17 g of Mg react, how many L of H 2 forms? Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) 17.0 g Mg 1 mol Mg 24.31 g Mg 1 mol H 2 1 mol Mg 22.4 L H 2 1 mol H 2 = 15.7 L H 2

20 b How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L Molar Volume Problems 2KClO 3  2KCl + 3O 2

21 63.55 g Cu 1 mol Cu Molar Volume Problems b How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M

22 MolarityMolarity b Molarity is the number of moles of solute dissolved in one liter of solution. b Units are moles per liter or moles of solute per liter of solution. b Molarity abbreviated by a capital M b Molarity = moles of solute liter of solution

23 Molarity Problems b As an example, suppose we dissolve 23 g of ammonium chloride (NH 4 Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution? 23 g NH 4 Cl 53.5 g NH 4 Cl 1 mole NH 4 Cl =.43 mol NH 4 Cl 145 mL 1000 mL 1 L =.145 L.43 mol NH 4 Cl.145 L = 2.97 M NH 4 Cl

24 Molarity Problems b Now, suppose we have a beaker with 175 mL of a 5.5 M HCl solution. How many moles of HCl is in this beaker? 1000 mL 1 L =.96 mol HCl 175 mL 1 L 5.5 mol HCl b Suppose you had 70 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution? 70 g NaCl 58.44 g NaCl 1 mol NaCl = 1.2 mol NaCl 1.2 mol NaCl 2.0 L = 0.6 M NaCl

25 Volume of Solutions b A 10.% HCl solution (soln) means: 10 g HCl (pure) 100 g HCl soln b A solution with a density of 1.5 g/mL means: 1.5 g soln 1 mL soln

26 Impure Substances b To say a substance is 75% NaCl by mass means: 75 g NaCl (pure) 100 g of NaCl solution or 100 g of NaCl solution 75 g of NaCL (pure) b Or, Iron ore that is 15% iron by mass means: 15 g Fe 100 g ore or 100 g of ore 100 g of ore 15 g of Fe 15 g of Fe

27 Energy & Stoichiometry

28 Exothermic and Endothermic b Exothermic process – heat is released into the surroundings Exo = Exit b Endothermic Process – heat is absorbed from the surroundings Endo = Into HEAT

29 Thermochemical Equations b In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product b Enthalpy: the heat content of a system at constant pressure (ΔH) b Endothermic (positive ΔH) 2NaHCO 3 + 129kJ Na 2 CO 3 + H 2 O + CO 2 b Exothermic (negative ΔH) CaO + H 2 O Ca(OH) 2 + 65.2kJ

30 Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ Fe(s) + O 2 (g)→ Fe 2 O 3 (s) + 1652 kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? 4 3 2 10.00g Fe 55.85g Fe 1 mol Fe 4 mol Fe 1652 kJ =73.97 kJ of heat Exo

31 Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = + 129 kJ: 2 NaHCO 3 + 129kJ → Na 2 CO 3 (s) + H 2 O + CO 2 How much heat is required to break down 50.0g of sodium bicarbonate? 50.0 g NaHCO 3 83.9 g NaHCO 3 1 mol NaHCO 3 2 mol NaHCO 3 129 kJ =38.4 kJ of heat Endo

32 Write the thermochemical equation for the synthesis of calcium oxide and water with a ΔH= - 65.2 kJ: CaO + H 2 O → Ca(OH) 2 + 65.2kJ How much energy is released when 100 g of calcium oxide reacts? 100 g CaO 56.07 g CaO 1 mol CaO 65.2 kJ =116 kJ of heat Exo

33 Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= + 61.5 kJ: 2 MgO + 61.5 → 2 Mg + O 2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy? 420 kJ 61.5 kJ 1 mol O 2 31.98 g O 2 =218 g of O 2 Endo

34 Stoichiometry in the Real World Stoichiometry – Ch. 11

35 Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

36 Limiting Reactants b Available Ingredients 24 graham cracker squares 1 bag of marshmallows 12 pieces of chocolate b Limiting Reactant chocolate b Excess Reactants Marshmallows and graham crackers

37 Limiting Reactants b Limiting Reactant one that is used up in a reaction determines the amount of product that can be produced b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

38 Limiting Reactant Steps 1. Write the balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product actually possible

39 Limiting Reactants b 79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed? Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 68.1 g

40 Limiting Reactants 2.02 g H 2 1 mol H 2 68.1 g HCl 1 mol HCl 36.46 g HCl = 1.89 g H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 68.1 g

41 Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 2.44 g H 2 1 mol H 2 1 mol Zn 2.02 g H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 68.1 g

42 Limiting Reactants Zn: 2.44 g H 2 HCl: 1.89 g H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 1.89 g H 2 left over zinc

43 Limiting Reactants #2 Limiting Reactants #2 b 5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed? 2Mg + O 2  2MgO 5.42 g ? g 4.00 g

44 Limiting Reactants #2 5.42 g Mg 1 mol Mg 24.31 g Mg = 8.99 g MgO 2 mol MgO 2 mol Mg 40.31 g MgO 1 mol MgO 2Mg + O 2  2MgO 5.42 g ? g 4.00 g

45 Limiting Reactants #2 40.31 g MgO 1 mol MgO 4.00 g O 2 1 mol O 2 32.00 g O 2 = 10.1 g MgO 2 mol MgO 1 mol O 2 2Mg + O 2  2MgO 5.42 g ? g 4.00 g

46 A. Limiting Reactants #2 Mg: 8.99 g MgOO 2 : 10.1 g MgO Excess oxygen Limiting reactant: Mg Excess reactant: O 2 Product Formed: 8.99 g MgO

47 Limiting Reactants b What other information could you find in these problems? How much of each reactant is used – in grams, liters, moles How much of excess reactant is left over – in grams, liters, moles

48 Percent Yield calculated on paper measured in lab

49 Percent Yield b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. 1 mol K 2 CO 3 138.2 g K 2 CO 3 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl 45.8 g K 2 CO 3 K 2 CO 3 + 2 HCl  2 KCl + H 2 CO 3 = 49.4 grams KCl

50 Percent Yield b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. 46.3 grams KCl 49.4 grams KCl x 100 % yield = 93.7 %


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