9.1 NOTES Stoichiometry

I. What is Stoichiometry? Definition: the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction; based on the law of conservation of mass and energy

Mole and mass relationships in chemical reactions balanced chemical equation gives the mole ratio coefficients represent not only individual particles, but also number of moles; mole ratio – ratio between the # of moles of any 2 substances # species in eqn. x next lower # = # of mole ratios (n x (n-1)) = # mole ratios;

i.e. Iron + Oxygen  Iron (III) oxide = 4Fe + 3O2  2Fe2O3 = 4 atoms of Fe + 3 molecules O2  2 formula units Fe2O3 4 moles Fe + 3 mole O2  2 moles Fe2O3 = 223.4g Fe + 96.0g O2  319.4g Fe2O3

II. Stoichiometric Calculations A. Mass-mass calculations Example: What mass of hydrogen gas can be formed when 12.5 grams of zinc reacts with excess hydrochloric acid?   How to approach a stoichiometry problem: 1. Read the problem, then write a balanced equation. 2. Identify the given and desired quantities. 3. Determine molar masses if necessary. 4. Carry out the factor-label calculation, labeling the answer with units.

a. Do the above example: Zn + 2HCl  ZnCl2 + H2 Given: 12.5 grams of Zn (65.4g/mol) Find: mass of H2 (2.02g/mol)   12.5gZn x 1 mol Zn x 1 mol H2 x 2.02 g H2 65.4 g Zn 1 mol Zn 1 mol H2 = 0.386 g H2

b. When 10.0 grams of propane (C3H8) is burned in air, what is the mass of water formed? C3H8 + 5O2  3CO2 + 4H2O   Given: 10.0g C3H8 *44.1g/mol) Find: mass of H2O (18.0g/mol) 10.0gC3H8 x 1 mol C3H8 x 4 mol H2O x 18.0g H2O 44.1 g C3H8 1 mol C3H8 1 mol H2O = 16.3g H2O