Stoichiometry Ch. 11. I. Stoichiometric Calculations

Transcription

1 Stoichiometry Ch. 11 I. Stoichiometric Calculations

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 Stoichiometry 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 1 bicycle = 37 bicycles 2 tires

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

5 Proportional Relationships Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio 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!! 4. Check answer.

7 Conversions LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass (g/mol) MOLES particles/mol NUMBER OF PARTICLES Molarity (mol/l) LITERS OF SOLUTION

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 e PbI 2

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

10 Mole to Mass We can also convert from moles to mass, and mass to moles For Example: 4 Al + 3 O 2 2Al 2 O 3

11 Mass to Moles: 4 Al + 3 O 2 2Al 2 O 3 If the reaction starts with.84 moles of aluminum, how many grams of aluminum oxide are produced?.84 mol Al 2 mol Al 2 O grams Al 2 O 3 4 mol Al Al2O3 = 42.8 grams Al 2 O 3

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

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

14 Stoichiometry with Gases 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. At STP e of gas occupies 22.4 L e = 22.4 L

15 Molar Volume In the following reaction, if 17 g of Mg react, how many L of H 2 forms? Mg (s) + HCl (aq) MgCl 2 (aq) + H 2 (g) 17.0 g Mg Mg H L H 2 = 15.7 L g H 2 Mg Mg H 2

16 Molar Volume Problems How many grams of KClO 3 are req d to produce 9.00 L of O 2 at STP? 2KClO 3 2KCl + 3O 2? g 9.00 L 9.00 L 2 mol O 2 O L O 2 KClO 3 3 mol O 2 g KClO 3 KClO 3 = 32.8 g KClO 3

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

18 Molarity Problems 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 e NH 4 Cl =.43 mol NH 4 Cl 145 ml 53.5 g NH 4 Cl 1 L 1000 ml =.145 L.43 mol NH 4 Cl.145 L = 2.97 M NH 4 Cl

19 Molarity Problems 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? 175 ml 5.5 mol HCl 1 L =.96 mol HCl 1000 ml 1 L

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

21 Energy & Stoichiometry

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

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

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

25 Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = kj: Endo 2 NaHCO kJ Na 2 CO 3 (s) + H 2 O + CO 2 How much heat is required to break down 50.0g of sodium bicarbonate? NaHCO g NaHCO g NaHCO3 129 kj 2 mol NaHCO 3 =38.4 kj of heat

26 Write the thermochemical equation for a single replacement of calcium oxide and water with a ΔH= kj: Exo CaO + H 2 O Ca(OH) kJ How much energy is released when 100 g of calcium oxide reacts? 100 g CaO CaO g CaO 65.2 kj CaO =116 kj of heat

27 Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= kj: Endo 2 MgO Mg + O 2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kj of Energy? 420 kj O g O 2 =218 g of O kj O2

28 Stoichiometry Ch. 11 Stoichiometry in the Real World

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

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

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

32 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

33 Limiting Reactants 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 g? g 68.1 g

34 Limiting Reactants Zn + 2HCl ZnCl 2 + H g? g 68.1 g g g Zn Zn H 2 H 2 = 2.44 g H 2 g Zn Zn H 2

35 Limiting Reactants Zn + 2HCl ZnCl 2 + H g? g 68.1 g g g HCl HCl H 2 H 2 = mol g H 2 g HCl HCl H 2

36 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

37 Limiting Reactants # 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 4.00 g? g

38 Limiting Reactants #2 2Mg + O 2 2MgO 5.42 g 4.00 g? g mol g g Mg Mg MgO MgO = 8.99 g mol MgO g Mg Mg MgO

39 Limiting Reactants #2 2Mg + O 2 2MgO 5.42 g 4.00 g? g mol g O 2 O g O 2 MgO O 2 g MgO = 10.1 g MgO MgO

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

41 Limiting Reactants 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

42 Percent Yield measured in lab actual yield % yield theoretical yield 100 calculated on paper

43 Percent Yield 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. K 2 CO HCl 2 KCl + H 2 CO g K 2 CO 3 K 2 CO g K 2 CO 3 2 mol KCl K 2 CO g KCl KCl = 49.4 grams KCl

44 Percent Yield 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 grams KCl 49.4 grams KCl x 100 % yield = 93.7 %