Chemical Equations A chemical equation just like a mathematical equation is a way to express, in symbolic form, the reactions occurring in a chemical system. n Balancing chemical equations n Reaction stoichiometry n Reagents limiting the extent of reaction n Acid-base reactions n Oxidation states of reactants and products n Combination Reactions: Atoms or molecules combine to form a new molecule 2H 2 + O 2 2H 2 O ClO + NO 2 ClNO 3 ClO + ClO ClOOCl n Decomposition Reactions: A molecule breaks apart to form different molecules or atoms NO 2 + sunlight O + NO 2C 7 H (NO 2 ) 3 + heat 7CO(g) + 7C + H 2 O(g) + 3N 2 (g) n Displacement Reactions: An atom or molecule displaces an atom or molecule in the reaction partner H + Cl 2 HCl + Cl 2Na(s) + 2H 2 O( ) 2NaOH(aq) + H 2 (g) n Exchange Reactions: The components of two compounds are exchanged BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) 3NaOH(aq) + H 3 PO 4 (aq) Na 3 PO 4 (aq) + 3H 2 O( ) Balancing Chemical Equations n Remember Conservation of Mass rule matter is neither created nor destroyed in a chemical reaction. n A balanced chemical equation must have the same number of each type of atom on the reactant side as on the product side. 1
Balancing Chemical Equations n Process for balancing chemical equations: 1. Determine correct chemical formulas of all reactants and products 2. Start with heavier atoms balance number of these on reactant and product sides of equation 3. If elements appear in equation as either reactants or products, balance these last 4. Electrical charge must be balanced Write a balanced chemical equation for the reaction of Fe with oxygen to form iron(iii) oxide 1. Fe + O 2 Fe 2 O 3 2. 2 Fe + O 2 Fe 2 O 3 Fe is currently balanced 3. 2 Fe + 3/2 O 2 Fe 2 O 3 O is now balanced coefficients must usually be integer numbers multiply everything by 2 to remove 3/2 denominator: 4 Fe + 3 O 2 2 Fe 2 O 3 Write a balanced chemical equation for the combustion of methane combustion is reaction with oxygen producing a flame. Complete combustion produces only CO 2 and H 2 O. 1. CH 4 + O 2 CO 2 + H 2 O 2a. CH 4 + O 2 CO 2 + H 2 O C is now balanced 2b. CH 4 + O 2 CO 2 + 2 H 2 O H is now balanced 3. CH 4 + 2 O 2 CO 2 + 2 H 2 O O is now balanced CH 4 + 2 O 2 CO 2 + 2 H 2 O Combustion of propane, C 3 H 8 1. C 3 H 8 + O 2 CO 2 + H 2 O 2a. C 3 H 8 + O 2 3 CO 2 + H 2 O 2b. C 3 H 8 + O 2 3 CO 2 + 4 H 2 O 3. C 3 H 8 + O 2 3 CO 2 + 4 H 2 O 4. No charges to balance C 3 H 8 + O 2 3 CO 2 + 4 H 2 O Reaction of ammonium perchlorate with aluminum NH 4 ClO 4 (s) + Al(s) Al 2 O 3 (s) + N 2 (g) + HCl(g) + H 2 O(g) 2a. Balance N: Al 2 O 3 (s) + N 2 (g) + HCl(g) + H 2 O(g) 2b. Balance Cl: Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + H 2 O(g) 2
Reaction of ammonium perchlorate with aluminum (con t.) 2c. Balance H: 2d. Balance O: Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + 3 H 2 O(g) /3 Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + 3 H 2 O(g) Reaction of ammonium perchlorate with aluminum (con t.) 3a. Balance Al: 2 NH 4 ClO 4 (s) + 10/3 Al(s) /3 Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + 3 H 2 O(g) 3b. Remove fractional coefficients (multiply by 3): 6 NH 4 ClO 4 (s) + 10 Al(s) Al 2 O 3 (s) + 3 N 2 (g) + 6 HCl(g) + 9 H 2 O(g) Mg(s) + H 3 O + (aq) Mg 2+ (aq) + H 2 (g) + H 2 O(l) 1. As written, the only element out of balance is the hydrogen balance H first: Mg(s) + 2 H 3 O + (aq) Mg 2+ (aq) + H 2 (g) + 2 H 2 O(l) 2. Check charge balance: left-hand side has total charge of +2 (2 * 1+ charge on hydronium ion) Right-hand side has total charge of +2 (2+ charge on Mg ion) of Chemical Reactions For the generic chemical reaction aa + bb + xx + yy + A, B, X, and Y represent the atoms or molecules reacting and forming, and the coefficients a, b, x, and y represent the stoichiometric coefficients they tell how many moles of one substance reacts with another substance to form some number of moles of the products. It is important to remember that the stoichiometric coefficients represent numbers of moles, not masses of reactants and products How much carbon dioxide is produced by burning one gallon of gasoline? n Gasoline is composed of many hydrocarbons, but let s assume they are all iso-octane (C 8 H 18 ) n Isooctane has a density of 0.6980 g ml -1 n 1 gal = 3.78 L = 378 ml 3
n Mass of isooctane (0.6980 g ml -1 ) (378 ml) = 2642 g C 8 H 18 n Molar mass: 8(12.011 g mol -1 ) + 18(1.0079 g mol -1 ) = 114.230 g mol -1 n Moles C 8 H 18 : n Balanced chemical equation: 2 C 8 H 18 + 2 O 2 16 CO 2 + 18 H 2 O n Determine moles of CO 2 produced: (23.13 mol C 8 H 18 ) (16 mol CO 2 ) (2 mol C 8 H 18 ) = 18.0 mol CO 2 (2642 g) / (114.230 g mol -1 ) = 23.13 mol C 8 H 18 n Determine mass of CO 2 : (18.0 mol CO 2 ) (44.009 g mol -1 ) = 8142 g CO 2 = 8.142 kg CO 2 per gallon of gasoline consumed n How much CO 2 is emitted by CA every year? Californians consume ~1.4 x 10 10 gallons of gasoline each year. (1.4 x 10 10 gal) (8.142 kg CO 2 gal -1 ) n The following general expression applies to stoichiometry problems: Moles B = (Coefficient B) (Coefficient A) (Moles A) = 1.1 x 10 11 kg CO 2 n Suppose the amount of one the reactant in a chemical reaction is insufficient to allow the reaction to proceed to completion. That reactant is called the limiting reagent. The limiting reagent limits how much product can formed in a reaction. 6 NH 4 ClO 4 (s) + 10 Al(s) n Al 2 O 3 (s) + 3 N 2 (g) + 6 HCl(g) + 9 H 2 O(g) If we begin with 1.00 kg ammonium perchlorate and 0.100 kg aluminum, how many moles of gaseous product will be produced? 1. Which reactant will be consumed first? (1000 g NH 4 ClO 4 )/(117.488 g mol -1 ) = 8.1 mol NH 4 ClO 4 (100 g Al)/(26.982 g mol -1 ) = 3.71 mol Al 4
1a. How much Al required for NH 4 ClO 4 to react completely? (10 mol Al) (8.1 mol NH 4 ClO 4 ) = 14.2 mol Al (6 mol NH 4 ClO 4 ) Since we only have 3.71 mol Al, there is insufficient Al for the NH 4 ClO 4 to react completely, so Al is the limiting reagent. 2. Determine moles of each gaseous product formed in reaction: (3 mol N 2 ) (3.71 mol Al) (10 mol Al) = 1.11 mol N 2 (6 mol HCl) (3.71 mol Al) = 2.23 mol HCl (10 mol Al) (9 mol H 2 O) (3.71 mol Al) = 3.34 mol H 2 O (10 mol Al) Total gas phase product = 6.68 mol Example Ammonia reacts with nitric oxide to form nitrogen gas and water: NH 3 + NO N 2 + H 2 O If 71.4 g NH 3 reacts with 168.6 g NO, how much N 2 and H 2 O will be produced? Step 1: Balance the chemical equation Example (con t.): NH 3 + NO N 2 + H 2 O Step 1: Balance the chemical equation 2 NH 3 + NO N 2 + 3 H 2 O H balanced 2 NH 3 + 3 NO N 2 + 3 H 2 O O balanced 2 NH 3 + 3 NO /2 N 2 + 3 H 2 O N balanced 4 NH 3 + 6 NO N 2 + 6 H 2 O remove fractional coefficient Example (con t.): NH 3 + NO N 2 + H 2 O Step 2: Determine moles of reactants 71.4 g NH3 = 4.19 mol NH3 17.031g/ mol 168.6 g NO =.62 mol NO 30.006 g/mol Step 3: Determine which is limiting reagent (4 mol NH3) (.62 mol NO) = 3.7 mol NH 3 NO is limiting (6 mol NO) reagent Example (con t.): NH 3 + NO N 2 + H 2 O Step 4: Determine amount of products ( mol N ) (.62 mol NO) 2 = 4.68 mol N (6 mol NO) 2 (4.68 mol N (.62 mol NO) 2) (28.014 g/mol) = 131g N2 (6 mol H2O) =.62 mol H2O (6 mol NO) (.62 mol H2O) (18.012 g/mol) = 101g H2O
Example Ca (PO 4 ) 3 F + H 2 SO 4 + 10 H 2 O compound initial mass initial moles Ca (PO 4 ) 3 F 1000. g 1.983 mol H 2 SO 4 200.0 g 2.039 mol H 2 O 100.0 g.1 mol Determine mass of all products formed Step 1: Determine limiting reagent examine Ca (PO 4 ) 3 F ( mol H SO ) (1.983 mol Ca...) 2 4 = 9.91 mol H2SO4 for complete rxn (1mol Ca...) (10 mol H O) (1.983 mol Ca...) 2 = 19.83 mol H2O for complete rxn (1mol Ca...) Not enough H 2 SO 4 or H 2 O to react completely with Ca (PO 4 ) 3 F Ca (PO 4 ) 3 F is not limiting reagent Step 1: Determine limiting reagent examine H 2 SO 4 (10 mol H O) (2.039 mol H SO ) 2 2 4 = 4.078 mol H2O for complete rxn ( mol H2SO4) H 2 O is in excess relative to the amount needed to react completely with H 2 SO 4 H 2 SO 4 is limiting reagent Step 2: Determine amount of product formed (3 mol H3PO4) (2.039 mol H2SO4) = 1.223 mol H3PO4 ( mol H2SO4) (1.223 mol H3PO4) ( 97.99 g/mol) = 119.8 g H3PO4 Step 2: Determine amount of product formed ( mol CaSO 2H O) (2.039 mol H SO ) 4 2 2 4 = 2.039 mol CaSO4 ( mol H2SO4) (2.039 mol CaSO4 2H2O) (172.17 g/mol) = 31.1g CaSO4 2H2O (1mol HF) (2.039 mol H2SO4) = 0.4078 mol HF ( mol H2SO4) (0.4078 mol HF) (20.006 g/mol) = 8.18 g HF Product Yields For each kg of Ca (PO 4 ) 3 F that reacts, 400. g of phosphoric acid are formed. What is the percent yield of phosphoric acid? 6
Product Yields Step 1: Determine the theoretical yield amount of product formed if reaction occurred completely (3 mol H PO ) (1.983 mol Ca...) 3 4 =.949 mol H3PO4 for complete rxn (1mol Ca...) (.949 mol H3PO4) (97.99 g/mol) = 83 g H3PO4 theoretical yield actual yield 400.g % yield = x 100% = x 100% = 68.6 % theor. yield 83g Example Problem 4.69 Cu ore contains Cu 2 S and CuS + 10% inert impurities. 200.0 g of ore is heated to produce 10.8 g of 90.0% pure Cu(s) and SO 2 (g). Find % Cu 2 S in ore. Cu 2 S + O 2 2 Cu(s) + SO 2 (g) CuS + O 2 Cu(s) + SO 2 (g) Step I: Determine moles of Cu formed (.900)(10.8 g impure Cu)/(63.46 g/mol) = 2.14 mol Cu Step II: Determine mass of Cu-containing ore (.900)(200.0 g ore) = 180.0 g Cu 2 S + CuS Example Problem 4.69 Cu ore contains Cu 2 S and CuS + 10% inert impurities. 200.0 g of ore is heated to produce 10.8 g of 90.0% pure Cu(s) and SO 2 (g). Find % Cu 2 S in ore. Step III: Set up equation for moles Cu in ore Let X = mass of Cu 2 S Let Y = mass of CuS X + Y = 180.0 g moles Cu in Cu 2 S and CuS: X g 2 mol Cu moles Cu from Cu2S = 19.17 g/mol 1 mol Cu2S moles Cu from CuS = Y g 1 mol Cu 9.611 g/mol 1 mol CuS Example Problem 4.69 Cu ore contains Cu 2 S and CuS + 10% inert impurities. 200.0 g of ore is heated to produce 10.8 g of 90.0% pure Cu(s) and SO 2 (g). Find % Cu 2 S in ore. Moles Cu in product = moles Cu in Cu 2 S & CuS 2X mol Y mol 2.14 mol = + Y = 180.0 g - X 19.17 9.611 2X mol (180.0 g - X) mol 2.14 mol = + 19.17 9.611 X = 122 g Y = 8 g 122 g % Cu2 S = 100% = 67.8 % 180 g 7