REVIEW STOICHIOMETRY STOICHIOMETRY What is it Demonstrates the

REVIEW STOICHIOMETRY

STOICHIOMETRY • What is it? • Demonstrates the relationships (ratios) between the amount of molecules used and produced in a chemical reaction • Helps us to determine how much of each reactant and/or product we will need/make • It is based on the Law of Conservation of Matter

STOICHIOMETRY • Steps to Follow • 1) Complete and balance the equation • 2) Set up your table (moles, molar mass, mass) • 3) Complete the table with the information your are given • 4) Determine what information you need to find • 5) Use molar mass to convert between moles and mass • 6) Use stoichiometric ratios to determine the number of moles

STOICHIOMETRY Combustion! • Example: If 60. 0 g of propane (C 3 H 8) is burned in a stove, carbon dioxide (CO 2) and water (H 2 O) are produced, what mass of CO 2 will be released into the environment? Step 1 – Complete and Balance the equation C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O

STOICHIOMETRY • Example: Step 2 – Set-up Table moles MM mass C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O

STOICHIOMETRY • Example: Steps 3 & 4 – Info given and needed moles C 3 H 8 + 5 O 2 → 3 CO 2 MM mass 60. 0 g ? g + 4 H 2 O

STOICHIOMETRY • Example: Steps 5 & 6 – Use ratios and molar mass to find the rest C 3 H 8 + moles 1. 36 mol MM 5 O 2 → 3 CO 2 44. 097 g/mol mass 60. 0 g ? g + 4 H 2 O

STOICHIOMETRY • Example: Steps 5 & 6 – Use ratios and molar mass to find the rest C 3 H 8 + moles 1. 36 mol MM 44. 097 g/mol mass 60. 0 g 5 O 2 → 3 CO 2 + 4. 08 mol 44. 009 g/mol ? g 4 H 2 O

STOICHIOMETRY • Example: Steps 5 & 6 – Use ratios and molar mass to find the rest C 3 H 8 + moles 1. 36 mol MM 44. 097 g/mol mass 60. 0 g 5 O 2 → 3 CO 2 + 4. 08 mol 44. 009 g/mol 180 ? g g 4 H 2 O