Unit 8 Stoichiometry Notes Balancing Equations, Stoichiometry, Limiting Reactants, % Yield

Stoichiometry Island Diagram Known Unknown Substance A Substance B M Mass Mass Mountain Mass Mole Island Volume Mole Mole Volume V Liter Lagoon Particles Particles P Particle Place Stoichiometry Island Diagram

Stoichiometry Island Diagram Known Unknown Substance A Substance B Mass Mass 1 mole = molar mass (g) 1 mole = molar mass (g) Use coefficients from balanced chemical equation Volume Mole Mole Volume 1 mole = 22.4 L @ STP 1 mole = 22.4 L @ STP (gases) (gases) 1 mole = 6.022 x 1023 particles (atoms or molecules) 1 mole = 6.022 x 1023 particles (atoms or molecules) Particles Particles Stoichiometry Island Diagram

Visualizing a Chemical Reaction 2 Na + Cl2 NaCl 2 ___ mole Na 10 10 ___ mole Cl2 5 5 ___ mole NaCl 10 10 ?

Visualizing a Chemical Reaction 2 Na + Cl2 NaCl 2 ___ mole Na 10 ___ mole Cl2 5 ___ mole NaCl 10

Formation of Ammonia N2 (g) + 3 H2 (g) 2 NH3 (g) + + + + + + + + 2 atoms N and 2 atoms N + 6 atoms H 6 atoms H 1 molecule N2 + 3 molecules H2 2 molecules NH3 10 molecule N2 + 30 molecules H2 20 molecules NH3 6.02 x 1023 molecules N2 6.02 x 1023 molecules H2 6.02 x 1023 molecules NH3 1 x + 3 x 2 x 1 mol N2 + 3 mol H2 2 mol NH3 28 g N2 + 3 x 2 g H2 2 x 17 g NH3 34 g reactants 34 g products Assume STP 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L + 22.4 L N2 67.2 L H2 44.8 L NH3

Standard Temperature & Pressure Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Molar Volume at STP MOLES LITERS OF GAS AT STP (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 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Mole  Mole Stoichiometric Calculation (GIVEN AMOUNT) (Mole Ratio from BALANCED Chemical Reaction) Given moles Unknown mole ratio Known mole ratio 2 : 1 : 2 Given 3 moles of O2. How many moles of H2O produced? 2H2 + O2  2H2O 3 moles O2 2 moles H2O 1 mole O2 = 6 moles H2O

Mole  Mass Stoichiometric Calculation (GIVEN AMOUNT) (Molar Mass) (Mole Ratio from BALANCED Chemical Reaction) Given mass 1 mol known Unknown mole ratio mass known Known mole ratio 2 : 1 : 2 Given 4 moles of H2. How many grams of H2O produced? 2H2 + O2  2H2O 4 moles H2 2 moles H2O 18.02 g H2O 2 mole H2 1 mol H2O = 72.08 grams H2O

Mass  Mass Stoichiometric Calculation (GIVEN AMOUNT) (Molar Mass Known) (Mole Ratio from BALANCED Chemical Reaction) (Molar Mass Unknown) Given mass 1 mol known Unknown mole ratio mass unknown mass known Known mole ratio 1 mol unknown 2 : 1 : 2 How many grams of H2 are needed to react with 64 grams of O2? 2H2 + O2  2H2O 64g O2 1 mole O2 2 mol H2 2.02g H2 32g O2 1 mol O2 1 mol H2 = 8.08 grams H2

Mole  Liters Stoichiometric Calculation (GIVEN AMOUNT) (Mole Ratio from BALANCED Chemical Reaction) Liter/mol conversion Given moles Unknown mole ratio 22.4 L unknown Known mole ratio 1 mol unkown 2 : 1 : 2 Given 3 moles of O2. How many liters of H2 needed? 2H2 + O2  2H2O 3 moles O2 2 moles H2 22.4 L H2 1 mole O2 1 mol H2 = 134.4 Liters H2

Stoichiometry Practice Problems 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 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

2KClO3  2KCl + 3O2 9 mol O2 2 mol KClO3 3 mol O2 = 6 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 2 mol KClO3 6 mol x mol KClO3 = 9 mol O2 = 6 mol KClO3 3 mol O2 O2 KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Stoichiometry Problems How many grams of KClO3 are required 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 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Stoichiometry Problems How many grams of KClO3 are required to produce 9.00 L of O2 at STP? 2KClO3  2KCl + 3O2 ? g 9.00 L 1 mol O2 2 mol KClO3 122.55 g KClO3 x g KClO3 = 9.00 L O2 = 32.8 g KClO3 22.4 L O2 3 mol O2 1 mol KClO3 32.8 g O2 KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

How many grams of KClO3 are required 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 32.8 g 1 mol O2 2 mol KClO3 122.55 g KClO3 x g KClO3 = 9.00 L O2 = 32.8 g KClO3 22.4 L O2 3 mol O2 1 mol KClO3 O2 KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

How many grams of silver will be formed from 12.0 g copper? Cu + 2 AgNO3  2 Ag + Cu(NO3)2 ? g 12.0 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 40.7 g 1 mol Cu 2 mol Ag 107.87 g Ag x g Ag = 12.0 g Cu = 40.7 g Ag 63.55 g Cu 1 mol Cu 1 mol Ag Cu Ag Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Calculating Limiting Reactants STEPS: Balance chemical reaction Choose a reactant and calculate the amount of product. Choose the other reactant and calculate the amount of product. Whichever reactant makes LESS product, that is your limiting reactant. The LEAST amount of product produced from Steps 3 and 4, is the amount theoretically produced by the reaction.