Unit 8: Stoichiometry: Part 1

Stoichiometry Means “element measuring” The study of quantitative relationships between the amounts of reactants used and the products formed by a chemical reaction Based on the law of conservation of mass

Mole Ratio In a balanced equation, the ratio between the numbers of moles of any 2 substances Indicated by the coefficients

2 Mg + O2  2 MgO 2 mol Mg 1 mol O2 Mg : O2 1 mol O2 2 mol Mg 2 mol Mg or 1 mol O2 2 mol Mg 2 mol Mg 2 mol MgO Mg : MgO or 2 mol MgO 2 mol Mg 1 mol O2 2 mol MgO MgO : O2 or 1 mol O2 2 mol MgO

Stoichiometry Steps 1. Write a balanced equation. 2. Identify the given & unknown. 3. Draw a roadmap 4. Set up railroad tracks 5. Calculate

Mole Ratio: Use coefficients from equation Mole  Mole Mole to Mole Ratio ___ mol Unknown mol Given ___ mol Given Mole Ratio: Use coefficients from equation

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

Mole  Mass ___ mol Unknown mol Given ___ mol Given 1 mol Unknown Molar Mass from Periodic table Mole to Mole Ratio ___ mol Unknown Molar mass (g) of Unknown mol Given ___ mol Given 1 mol Unknown Mole Ratio: Use coefficients from equation

mol O2  mol of H2O  grams of H2O Ex: How many grams of H2O would be required to produce 5 moles of O2? 2H2O  2H2 + O2 ? grams 5 mol mol O2  mol of H2O  grams of H2O 5 mol O2 2 mol H2O 1 mol O2 18.02 g H2O 1 mol = 180.2 g H2O

Mole Ratio: Use coefficients from equation Mass  Mole Mole to Mole Ratio ___ mol Unknown g Given 1 mol Given Molar mass (g) of Given ___ mol Given Mole Ratio: Use coefficients from equation Molar Mass from Periodic table

grams NH3  mol NH3  mol NO 4NH3 + 5O2  4NO + 6H2O 824 g NH3 1 mol Ex: How many moles of NO would be formed with 824 g of NH3? 4NH3 + 5O2  4NO + 6H2O 824 g ? mol grams NH3  mol NH3  mol NO 824 g NH3 1 mol NH3 17.04 g NH3 4 mol NO NH3 = 48.4 mol NO

Mass  Mass ___ mol Unknown g Given 1 mol Given Molar Mass from Periodic table Mole to Mole Ratio ___ mol Unknown Molar mass (g) of Unknown g Given 1 mol Given Molar mass (g) of Given ___ mol Given 1 mol Unknown Mole Ratio: Use coefficients from equation Molar Mass from Periodic table

g Cu  mol Cu  mol Ag  g Ag Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g Ex: How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g g Cu  mol Cu  mol Ag  g Ag 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

Unit 8: Stoichiometry: Part 2

How many sandwiches can you make? Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly What limits the amount? bread What is left over? peanut butter and jelly

Limiting vs. Excess Excess Reactant Limiting Reactant The reactant that used up first in a reaction Determines the amount of product Excess Reactant The reactant that is left over after the reaction stops

Percent Yield The ratio of actual yield (from an experiment) to theoretical yield (from stoichiometric calculations) expressed as a percent

Percent Yield Actual Yield: The amount of product actually produced when a reaction is carried out in an experiment Theoretical Yield: The maximum amount of product that can be produced from a given amount of reactant

Percent Yield Equation measured in lab calculated on paper

When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed 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 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g Theoretical Yield: 45.8 g K2CO3 1 mol K2CO3 138.21 g 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g actual: 46.3 g Theoretical Yield = 49.4 g KCl 46.3 g 49.4 g % Yield =  100 = 93.7%