Chemistry 2015-10-22 chapter 8 notes PDF

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1 CHAPTER 8: STOICHIOMETRY Properties of substances  Physical property—property that a substance displays without a change in its makeup or composition o Odor, taste, color, appearance, melting point, boiling point, etc.  Physical change—change of state of a substance o Solid → liquid → gas o Not chemical  Chemical property—property that a substance displays as it alters its composition through a chemical change o Corrosiveness, acidity, flammability, combustibility, and reactivity with other substances o Example: group 1A metals will react with halogens to form ionic compounds  Chemical change—atoms rearrange to transform the original atoms or molecules into new and different substances Writing and balancing equations  Step 1: write correct formulas (skeletal equation for each reactant and product as outlined in chapter 5  Step 2: Balance atoms that occur in more complex molecules first, then balance atoms in pure elements (or diatomic molecules) last o Number of each different atom should be equal on both sides of the arrow  Step 3: always balance atoms in free elements, diatomic molecules, or any compound by adjusting only the coefficient in front of each species o NEVER change a subscript in a formula in order to balance atoms  Step 4: if balanced equation contains fractional coefficients such as 1/2 or 2/3, simplify these by multiplying ALL coefficients by the denominator in that fraction  Step 5: recheck balancing efforts by summing total number of each type of atom on both sides of equation EXAMPLE: balancing equations CH4 + 2O2 → CO2 + 2H2O C3H6 + xO2 → 3CO2 + 3H2O 2x coefficient of O2 = 9 9 → multiply ALL coefficients by 2 x= 2 2C3H6 + 9O2 → 6CO2 + 6H2O Reaction stoichiometry  Mass and molar relationships that exist in a chemical reaction  Mole-mole ratio—relates number of moles o substance B to number of moles of substance A 1 mol A 1 mol B gB o gAx = mol A x = mol B x =gB xgA 2 mol A 1 mol B EXAMPLE: mass calculation Calculate mass (g) of CO2 produced by combustion of 3.5 x 1015 g octane in excess O2 2C8H18(l) + 25O2(g) → 16CO2 + 18H2O(g) 1 mol octane 16 mol CO 2 44.01 g x x = 1.1 x 1016 g CO2 3.5 x 1015 g octane x 114.26 g octane 2 mol octane 1 mol CO 2

2 Limiting reactant and yield  Limiting reactant—limits how much product will form and is consumed first o Theoretical yield—limiting reactant will produce smallest amount of product given mole-mole relationships in balanced reaction  Yield—how much product is actually formed vs theoretical yield based on ratios in stoichiometric equation o Actual yield—actual amount of product isolated and weighed (in grams) at the end of a reaction actual yield x 100%  Percent yield— theoretical yield EXAMPLE: limiting reactant If we have 5 moles of CH4 and 8 moles of O2, which is the limiting reactant? How many moles of carbon dioxide will form? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) 1 mol CO 2 = 5 mol CO2 5 mol CH4 x 1 mol CH 4 1 mol CO 2 = 4 mol CO2 8 mol O2 x 2mol O 2 Limiting reactant—O2 because 4 < 5 2.5 Percent yield = x 100% = 62.5% 4 EXAMPLE: calculating theoretical amount What is the theoretical amount of AlCl3 that would form? 2Al(s) + 3Cl2(g) → 2AlCl3(s) Added together: 2.0 g Al + 2.0 g Cl2 1 mol Al 2 mol AlCl3 2.0 g Al x x = 0.074 mol AlCl3 26.98 g Al 2 mol Al 1 mol Cl 2 2 mol AlCl3 2.0 g Cl2 x x = 0.0188 mol Cl2 70.90 g Cl 2 3 mol Cl 2 Limiting reactant—Cl2 133.33 g = 2.5 g AlCl3 Theoretical yield—0.0188 mol AlCl3 x 1 mol AlCl3 Cl (35.45)3 + Al (26.98) = 133.33 g/mol EXAMPLE: chapter 8 question 45 2Na(s) + Br2(g) → 2NaBr(s) 2 mol NaBr 2 mol Na x = 2 mol NaBr 2 mol Na 2 mol NaBr 2 mol Br2 x = 4 mol NaBr 1 mol Br 2 Limiting reactant—Na because 2 < 4 EXAMPLE: mole calculations How many mol O2 are required to react completely with 7.2 mol C6H14? 19 O2(g) → 6CO2(g) + 7H2O(g) C6H14(g) + 2 2C6H14 + 19O2 → 12CO2 + 14H2O

3 19 mol O 2 7.2 mol C6H14 x 2 mol C 6 H 14

= 68 mol O2

EXAMPLE: moles formed How many moles NH3 forms when 2.6 mol N2H4 completely reacts? 3N2H4(l) → 4NH3(g) + N2(g) 4 mol NH 3 2.6 mol N2H4 x = 3.5 mol NH3 3 mol N 2 H 4 3 mol N 2 H 4 = 7.8 mol N2H4 2.6 mol N2 x 1mol N 2 Types of chemical reactions  Combustion—reaction of a substance with oxygen to produce an oxygen-containing compound  Double replacement— AB + CD → AD + CB o Acid base reactions, precipitation reactions where partners are exchanged, etc.  HCl(aq) + NaOH(aq) → NaCl(aq) + H2O o Precipitation reactions—solid precipitate is formed  Cu+2(aq) + 2OH- → Cu(OH)2(s)  (s)—insoluble  (aq)—aqueous o Soluble and an be separated into its component ions o Gas evolving reactions— K2S(aq) + H2SO4(aq) → K2SO4(aq) + H2S(g)  Gas formed directly from ion exchange or by decomposition of one of products into gas and water  Single replacement— AB + C → CB + A o Oxidation (redox) reaction—one partner is exchanged for another as one species is oxidized and the other is reduced  Decomposition reaction— AB → A + B  Alkali metal—can react with halogens to produce metal halides o React with water to form metal ions, hydroxide ion (OH-) and hydrogen gas  Halogen—all react with many metals to form metal halides containing ionic bonds o React with hydrogen to form hydrogen halides containing covalent bonds o React with each other to form interhalogen compounds containing covalent bonds...