Chapter 12: Stoichiometry Table of Contents Chapter 12: Stoichiometry 12.1: What is Stoichiometry? 12.2: Stoichiometric Calculations

DEMOS!  Combustion of Methane CH4 + O2 → CO2 + H2O 1 CH4 : 1 O2 8 ml CH4 : 8 ml O2 16 ml total CH4 + 2O2 → CO2 + 2H2O 1 CH4 : 2 O2 8 ml CH4 : 16 ml O2 24 ml total 1st: FLAME: Add oxygen 1st – always have a little more fuel (methane) the 1st time around so that it doesn’t go off 2nd: BOOM: Add CH4 1st – mixes in better – more oxygen is best

DEMOS!  Combustion of Propane C3H5 + O2 → CO2 + H2O 1 C3H5 : 2 O2 3 ml C3H5 : 6 ml O2 9 ml total C3H5 + 5O2 → 3CO2 + 4H2O 1 C3H5 : 5 O2 3 ml C3H5 : 15 ml O2 18 ml total 1st: FLAME: Add oxygen 1st – always have a little more fuel (methane) the 1st time around so that it doesn’t go off 2nd: BOOM: Add CH4 1st – mixes in better – more oxygen is best

DEMOS!  Combustion of Butane C4H10 + O2 → CO2 + H2O 2 C3H5 : 13 O2 2 ml C3H5 : 13 ml O2 15 ml total 1st: FLAME: Add oxygen 1st – always have a little more fuel (methane) the 1st time around so that it doesn’t go off 2nd: BOOM: Add CH4 1st – mixes in better – more oxygen is best

Chapter 12: Stoichiometry Table of Contents Chapter 12: Stoichiometry Stoichiometry is the mathematical study of chemical reactions. 12.1: Mole-Mass Stoichiometry ratios The molar ratio of one reactant/product to another reactant/product 12.2: Stoichiometric Calculations Used to calculate how much reactant (or product) will be consumed (or created) in a chemical reaction. It shows quantitative relationship between the reactants and products.

2 H2 molecules + 1 O2 molecule → 2 H2O molecules Stoichiometry: Basic Concepts Stoichiometry The coefficients represent… (Ionic compounds = formula units) 2H2 + O2 → 2H2O 2 H2 molecules + 1 O2 molecule → 2 H2O molecules 2 moles H2 + 1 mole O2 → 2 moles H2O ratio: 2 mol H2 : 1 mol O2

Stoichiometry problems Stoichiometry: Basic Concepts Stoichiometry problems 1. Write balanced equation a. Identify reactants and products b. Write their chemical formulas (balance charges) c. Balance the equation

Stoichiometry problems Stoichiometry: Basic Concepts Stoichiometry problems 2. Convert to moles Just like we’ve been practicing 3. Convert from moles known to moles of unknown (from balanced equation) 4. Convert moles to needed units

Stoichiometry: Moles–Moles equations Ex1) How many moles sodium hydroxide can be produced when 3.12 moles sodium sulfate are reacted with excess potassium hydroxide (double displacement)? Na2SO4 + KOH --> K2SO4 + NaOH 2 2 3.12 mol Na2SO4 2 mol NaOH = 6.24 mol NaOH 1 mol Na2SO4

Stoichiometry: Moles–Moles equations Na2SO4 + 2KOH --> K2SO4 + 2NaOH Ex2) How many moles potassium sulfate can be produced from 7.36 moles potassium hydroxide? 7.36 mol KOH 1 mol K2SO4 = 3.68 mol K2SO4 2 mol KOH

Stoichiometry: Moles–Moles equations Na2SO4 + 2KOH --> K2SO4 + 2NaOH Ex 3) How many moles of potassium hydroxide are required to react with 11.47 moles sodium sulfate? 11.47 mol Na2SO4 2 mol KOH = 22.94 mol Na2SO4 1 mol Na2SO4

Stoichiometry: Moles–Moles equations Na2SO4 + 2KOH --> K2SO4 + 2NaOH Ex 4) How many moles of potassium hydroxide did you use to create 7.12 moles NaOH? 7.12 mol NaOH 2 mol KOH = 7.12 mol KOH 2 mol NaOH

Just so you know… WA ch 12 + 11.4, 11.5 is ENORMOUS (200 points) You can now do #11-18 Answers for #7-9 are given on the website mmecall.weebly.com You can also find… Notes homework hints