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9.1 NOTES Stoichiometry.

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Presentation on theme: "9.1 NOTES Stoichiometry."— Presentation transcript:

1 9.1 NOTES Stoichiometry

2 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

3 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;

4 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 g O2  319.4g Fe2O3

5 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.

6 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 g H2 65.4 g Zn mol Zn mol H2 = g H2

7 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 g H2O 44.1 g C3H mol C3H mol H2O = 16.3g H2O

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