Presentation on theme: "Stoichiometry: The study of quantitative measurements in chemical formulas and reactions Chemistry – Mrs. Cameron."— Presentation transcript:
Stoichiometry: The study of quantitative measurements in chemical formulas and reactions Chemistry – Mrs. Cameron
Review of Chemical Equations The Law of Conservation of Mass applies to all chemical equations. All equations must be balanced to have the same number and type of atoms on both the product and reactant sides of an equation. Equations are balanced by adding coefficients in front of compound formulas in chemical equations.
Consider the reaction of ethylene with oxygen: C 2 H 4 + 3O 2 2CO 2 + 2H 2 O Can be read as: 1 molecule C 2 H molecules O 2 2 molecules CO molecules H 2 O OR: 1 mole C 2 H moles O 2 2 moles CO moles H 2 O
C 2 H 4 + 3O 2 2CO 2 + 2H 2 O The coefficients in a chemical equation provide the ratio in which moles or molecules of one substance react with moles or molecules of another. The coefficients in a chemical equation provide the ratio in which moles or molecules of reactants relate to the moles or molecules of the product(s).
C 2 H 4 + 3O 2 2CO 2 + 2H 2 O If the coefficients represent moles rather than molecules, everything in the equation is enlarged by a factor of Avogadros #. For example: Every 3 moles of O 2 require 1 mole of C 2 H 4 Every 3 moles of O 2 produce 2 moles of CO 2
C 2 H 4 + 3O 2 2CO 2 + 2H 2 O The Coefficients provide the relational quantities of reactant(s) and product(s). Within an equation, they can be used to form a molar ratio between any two substances in the reaction. Every 3 moles of O 2 produce 2 moles of CO 2 OR
C 2 H 4 + 3O 2 2CO 2 + 2H 2 O Stoichiometry uses larger or smaller quantities – like cutting recipe in half or doubling or tripling a recipe Stoichiometry allows for calculations of multiples of the standard equations.
(63.4 g O 2 ) = 58.1 g CO 2 Molar Ratio
Calculating volume of a gas from a mass of another reactant or product.
Simplification of Volume-Volume Problems
Applications of Stoichiometry Problems Limiting Reactants –The limiting reactant limits the amount of product that can be formed –It is related to the molar ratio like the measurements in a recipe –The limiting reactant is totally consumed or used up in a chemical reaction Excess Reactants - the reactant left over.
N 2 (g) + 3H 2 (g) 2NH 3 (g) In this reaction, nitrogen reacts with hydrogen to form ammonia. For every one molecule (or mole) of nitrogen, you need 3 molecules (or moles) of hydrogen. Since there are only enough hydrogen molecules (3 sets of 3) to make 3 recipes of the ammonia, and not enough to make another, the hydrogen is the limiting reactant.
N 2 (g) + 3H 2 (g) 2NH 3 (g) In this reaction, nitrogen reacts with hydrogen to form ammonia. For every one molecule (or mole) of nitrogen, you need 3 molecules (or moles) of hydrogen. Since there are two molecules of nitrogen left over, it is the excess reactant.
Calculating Limiting Reactant Problems Balance the equation Complete a mass-mass problem for each reactant and the same product. Whichever reactant makes the least product is the limiting reactant. Whichever reactant makes the most product will have some leftover, or is the excess reactant.
Example: C 2 H 4 + 3O 2 2CO 2 + 2H 2 O Ethylene reacts with oxygen to produce carbon dioxide and water. If 8.0 g of ethylene and 16.3 g of oxygen react, how many grams of water can be produced? 8.00 g C 2 H g O 2 Since the oxygen produces less product, it is the limiting reactant. The ethylene is the excess reactant. From this basic calculation, a chemist could also calculate how much ethylene was left over. = 10.3 g H 2 O = 6.11 g H 2 O
Percent Yield Problems Theoretical Yield –The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. The actual yield (amount produced) of a reaction is usually less than the maximum expected (theoretical yield). Percent Yield –The actual amount of a given product as the percentage of the theoretical yield.
Percent Yield Problems In the reaction above, it was found that only 5.20 grams of water were produced. What is the percent yield of this reaction? X 100 % = 85.1% yield An example: