Stoichiometric Calculations (p. 275-287) Stoichiometry – Ch. 9 Stoichiometric Calculations (p. 275-287)

A. 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

A. 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 2 Mg + O2  2 MgO

Standard Temperature & Pressure B. Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm Note: When not at STP the ideal gas law will be used to convert between moles ↔ liters of gas (cover this later)

C. Avogadro’s Number Number of particles in one mole of a substance. 1 mole = 6.022 x 1023 particles Can be used as a conversion factor between moles and Number of representative particles. Reminder: The representative particles for an element is an atom for a molecular compound is a molecule for an ionic compound is a formula unit

D. Molarity Molarity is a unit of concentration – it tells us the amount of solute in one liter of solution Molarity can be used to convert between moles and volume of solution. Molarity, M = moles of solute / liters of solution or Molarity, M = moles of solute / 1000 ml of solution

E. Molar Mass Molar Mass serves as the conversion factor between grams and moles. It has the units of grams/1 mole grams → moles given grams x 1 mole / ? Grams moles → grams given moles x ? Grams / 1 mole ? Stands for gram-formula or gram – atomic mass of the substance.

F. Stoichiometry Steps Write a balanced equation. Label given and target. Convert given unit to moles using the appropriate conversion factors. Conversion Factors: moles ↔ grams Molar Mass moles ↔ number of particles Avogadro’s Number moles ↔ liters of solution Molarity moles ↔ liters of gas at STP Standard Molar Volume 4. Convert moles of given substance to moles of target substance using the mole ration from the balanced equation. Note: This is the core step in all stoichiometry problems! Convert moles of target to final unit using the appropriate conversion factor.

G. Connections between units LITERS OF GAS AT STP Molar Volume (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

H. Stoichiometry 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

Stoichiometry Problems 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

Stoichiometry Problems How many grams of silver will be formed from 12.0 g copper? 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 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

II. Stoichiometry in the Real World (p. 288-294) Stoichiometry – Ch. 9 II. Stoichiometry in the Real World (p. 288-294)

A. Limiting Reactants § Limiting Reactant bread § Excess Reactants § Available Ingredients 4 slices of bread 1 jar of peanut butter (50 oz) 1/2 jar of jelly (24 oz) § Recipe – 2 slices of bread, 1 oz jelly, 2 oz peanut butter. § Limiting Reactant bread § Excess Reactants peanut butter and jelly

A. Limiting Reactants Limiting Reactant used up in a reaction determines the amount of product Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

A. Limiting Reactants 1. Can be identified by the question involving amounts of more than one reactant. 2. For each reactant, calculate the amount of product formed using the 5 steps of stoichiometry. 3. Smaller answer indicates: limiting reactant amount of product

A. Limiting Reactants Reminder - The five steps of stoichiometry Write a balanced chemical equation. Label your given and target substances. Convert your given unit(s) to moles of given substance using the appropriate conversion factor. 4. Convert moles of given substance to moles of target substance using the mole ratio from the balanced equation. Convert moles of target substance to the final unit using the appropriate conversion factor. Note: Carry out steps 3-5 as needed for both reactants in a limiting reactant problem.

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 22.4 L H2 1 mol = 27.1 L H2

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 2.5 mol HCl 1 L 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2

A. Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 left over zinc

B. Percent Yield measured in lab calculated on paper

B. 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. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

B. Percent Yield 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

B. Percent Yield 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%