Stoichiometric Calculations Stoichiometry – Ch. 11 Stoichiometric Calculations

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

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

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? Ratio of eggs to cookies 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

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 + O2  2 MgO

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 Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

Conversions LITERS OF GAS AT STP Molar Volume MASS IN GRAMS MOLES (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

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(NO3)2 + 2KI  2 KNO3 + PbI2 Mole ratio of potasium iodide to lead (II) iodide: 2 moles KI 1 mole PbI2

Mole to Mole Problems 2KClO3  2KCl + 3O2 9 mol O2 2 mol KClO3 How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3  2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

Mole to Mass We can also convert from moles to mass, and mass to moles For Example: 4 Al + 3 O2  2Al2O3

Mass to Moles: 4 Al + 3 O2  2Al2O3 If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced? .84 mol Al 2 mol Al2O3 101.9 grams Al2O3 4 mol Al 1 mol Al2O3 = 42.8 grams Al2O3

Mass to Mass Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g Cu 1 mol Cu 63.55 How many grams of silver will be formed from 12.0 g copper reacting with silver nitrate? Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag

Stoichiometry Problems – Mole-Mole N2H4 + N2O4  N2 + H2O ? moles N2O4 = 2.72 moles N2H4 2 3 4 2.72 mol ? mol 2.72 mol N2 H4 1 mol N2O4 2 mol N2 H4 = 1.36 mol N2O4

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 1 mole of gas occupies 22.4 L 1 mole = 22.4 L

Molar Volume = 15.7 L H2 17.0 g Mg 1 mol Mg 24.31 g 1 mol H2 Mg In the following reaction, if 17 g of Mg react, how many L of H2 forms? Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g) 17.0 g Mg 1 mol Mg 24.31 g 1 mol H2 Mg 22.4 L H2 1 mol H2 = 15.7 L H2

Molar Volume Problems 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

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

Molarity Problems = .43 mol NH4Cl = .145 L = 2.97 M NH4Cl .145 L As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution? 23 g NH4Cl 1 mole NH4Cl = .43 mol NH4Cl 53.5 g NH4Cl 145 mL 1 L = .145 L 1000 mL .43 mol NH4Cl = 2.97 M NH4Cl .145 L

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 1 L 5.5 mol HCl = .96 mol HCl 1000 mL 1 L

Molar Volume Problems Cu + 2AgNO3  2Ag + Cu(NO3)2 1.5 L .10 mol AgNO3 How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu

Energy & Stoichiometry

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

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) 2NaHCO3 + 129kJ Na2CO3 + H2O + CO2 Exothermic (negative ΔH) CaO + H2O Ca(OH)2 + 65.2kJ

Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ Fe(s) + O2(g)→ Fe2O3(s) + 1652 kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? Exo 4 3 2 10.00g Fe 1 mol 1652 kJ =73.97 kJ of heat 55.85g Fe 4 mol Fe

Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = + 129 kJ: 2 NaHCO3 + 129kJ → Na2CO3(s) + H2O + CO2 How much heat is required to break down 50.0g of sodium bicarbonate? Endo 1 mol NaHCO3 50.0 g NaHCO3 129 kJ 83.9 g NaHCO3 2 mol NaHCO3 =38.4 kJ of heat

Write the thermochemical equation for a single replacement of calcium oxide and water with a ΔH= - 65.2 kJ: CaO + H2O → Ca(OH)2 + 65.2kJ How much energy is released when 100 g of calcium oxide reacts? Exo 100 g CaO 1 mol CaO 65.2 kJ 56.07 g CaO 1 mol CaO =116 kJ of heat

Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= + 61.5 kJ: 2 MgO + 61.5 → 2 Mg + O2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy? Endo 420 kJ 1 mol O2 31.98 g O2 =218 g of O2 61.5 kJ 1 mol O2

Stoichiometry in the Real World Stoichiometry – Ch. 11 Stoichiometry in the Real World

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

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

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

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

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? g 68.1 g 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  ZnCl2 + H2 79.1 g 68.1 g ? g

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 2.02 g H2 1 mol = 2.44 g H2

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 68.1 g HCl 1 mol HCl 36.46 g HCl 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 = 1.89 g H2

Limiting Reactants Zn: 2.44 g H2 HCl: 1.89 g H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 1.89 g H2 left over zinc

Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 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 + O2  2MgO 5.42 g 4.00 g ? g

Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 5.42 g Mg 1 mol Mg 24.31 g Mg 2 mol MgO Mg 40.31 g MgO 1 mol = 8.99 g MgO

Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 4.00 g O2 1 mol O2 32.00 g O2 2 mol MgO 1 mol O2 40.31 g MgO 1 mol MgO = 10.1 g MgO

A. Limiting Reactants #2 Limiting reactant: Mg Excess reactant: O2 Mg: 8.99 g MgO O2: 10.1 g MgO Limiting reactant: Mg Excess reactant: O2 Product Formed: 8.99 g MgO Excess oxygen

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

Percent Yield measured in lab calculated on paper

Percent Yield 2 mol KCl 74.55 g KCl 1 mol 1 mol 45.8 g K2CO3 K2CO3 When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2 HCl  2 KCl + H2CO3 1 mol K2CO3 138.2 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol 45.8 g K2CO3 = 49.4 grams KCl

46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 % Percent Yield When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. 46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 %