# Stoichiometry and Aqueous Reactions (Chapter 4)

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1 Stoichiometry and Aqueous Reactions (Chapter 4) Chemical Equations 1. Balancing Chemical Equations (from Chapter 3) Adjust coefficients to get equal numbers of each kind of element on both sides of arrow. Use smallest, whole number coefficients. e.g., start with unbalanced equation (for the combustion of butane): C4H10 + O2 CO2 + H2O reactants products { Hint -- first look for an element that appears only once on each side; e.g., C } C 4 H /2 O 2 4 CO H 2 O multiply through by 2 to remove fractional coefficient: 2 C 4 H O 2 8 CO H 2 O 2. Reaction Stoichiometry -- Mole Method Calculations Coefficients in balanced equation give the ratio by moles!!! e.g., in the above reaction: 2 moles C 4 H 10 react with 13 moles O 2 to produce 8 moles CO2 and 10 moles H2O Use these just like other conversion factors! How many moles of O 2 are required to react with 0.50 moles of C 4 H 10 according to the above equation? 0.50 mole C 4 H 10 x 13 mole O 2 = 3.25 mole O 2 2 mole C4 H 10 Page 1

2 Always convert given quantities (e.g., grams) to Moles!!! grams A moles A moles B grams B What mass of CO2 could be produced from the combustion of 100 grams of butane (C 4 H 10 )? will need formula masses to convert between grams and moles: CO2 = g/mole C4H10 = g/mole Step 1: "grams A moles A" 100 g C 4 H 10 x 1 mole C 4 H 10 = mole C 4 H g C 4 H10 Step 2: "moles A moles B" mole C 4 H 10 x 8 mole CO 2 = mole CO 2 2 mole C4H10 Step 3: "moles B grams B" mole CO2 x g CO2 = 303 g CO2 1 mole CO 2 Alternatively, the 3 steps can be combined into one Factor-Label string: 100 g C4H10 x 1 mole C4H10 x 8 mole CO g C4H10 2 mole C4H10 x g CO 2 = 303 g CO 2 1 mole CO2 Page 2

3 3. Limiting Reactant Calculations In practice, reactants are often combined in a ratio that is different from that in the balanced chemical equation. One of the reactants will be completely consumed and some of the other will remain unreacted. The limiting reactant is the one that is completely consumed. It determines the maximum amount (yield) of the products. Whenever quantities of both reactants are given, the limiting reactant must be determined!!! In a commercial process, nitric oxide (NO) is produced as follows: 4 NH O 2 4 NO + 6 H 2 O What mass (in grams) of NO can be made from the reaction of g NH 3 and g O 2? 1st, find moles of reactants: g NH 3 x 1 mole NH 3 = mole NH g NH g O 2 x 1 mole O 2 = mole O g O 2 2nd, calculate amount of product based on each reactant, separately: yield of NO based on NH3: mole NH 3 x 4 mole NO = mole NO 4 mole NH 3 yield of NO based on O2: mole O 2 x 4 mole NO = mole NO 5 mole O2 Therefore, O2 is the limiting reactant! (excess of NH3 exists) 3rd, calculate yield of product based on limiting reactant: mole NO x g NO = g NO 1 mole NO Page 3

4 4. Theoretical and Percentage Yield In actual experiments, the amount of a product that is actually obtained is always somewhat less than that predicted by the stoichiometry of the balanced chemical equations. This is due to competing reactions and/or mechanical losses in isolation of the product. actual yield -- amount of product obtained experimentally theoretical yield -- amount of product predicted by balanced equation percentage yield = (actual yield / theoretical yield) x 100% In the previous experiment for the production of NO from g NH 3 and g O 2, the chemist obtained g NO. What is the percentage yield of this reaction? theoretical yield = g NO (based on limiting reactant as above) actual yield = g (given in problem) % yield = (25.50 g NO / g NO) x 100% = 85.0 % { Note: the % yield can never be more than 100 % } Page 4

5 Solution Concentrations and Solution Stoichiometry 1. Solution Terminology solution - homogeneous (uniform) mixture, consisting of: solvent - the bulk medium, e.g., H 2 O solute(s) - the dissolved substance(s), e.g., NaCl concentration - measure of relative solute/solvent ratio standard solution - accurately known concentration saturated solution - contains maximum amount of solute solubility - concentration of a saturated solution, e.g.: solubility of NaCl is about 36 g NaCl / 100 g H 2 O ("soluble") solubility of CuS is about 10-5 g CuS / 100 g H2O ("insoluble") precipitate - an "insoluble" reaction product e.g., a precipitation reaction where the precipitate is AgCl: NaCl (aq) + AgNO 3 (aq) AgCl (s) + NaNO 3 (aq) 2. Molar Concentration Molarity (M) = moles solute / liter of solution units: moles/l or moles/1000 ml (just a conversion factor!) e.g., a "0.10 M" NaCl solution contains 0.10 mole NaCl per liter of solution What mass of NaCl is required to prepare 300 ml of 0.10 M solution? 1st - find moles of NaCl required: 300 ml x moles NaCl = moles NaCl 1000 ml 2nd - convert to grams of NaCl: moles NaCl x g NaCl = 1.75 g NaCl 1 mole NaCl Prepare this solution by weighing 1.75 g NaCl, dissolving in some H 2 O (about 250 ml), and then diluting to the 300 ml mark. Page 5

6 3. Dilution of Concentrated Solutions concentrated solution + H 2 O dilute solution (moles solute)conc = (moles solute) dil Vc M c = V d M d A 5.00 M NaCl "stock" solution is available. How would prepare 300 ml of a M NaCl "standard" solution? Vc x (5.00 M) = (300 ml) x (0.100 M) Vc = (300 ml) x (0.100 M) / (5.00 M) = 6.00 ml Measure out 6.00 ml of the 5.00 M "stock" solution, then add H 2 O to a total volume of 300 ml. 4. Stoichiometry Problems -- Reactions in Solution Start with a Balanced Equation and Use the Mole Method (Molarity is just a conversion factor!) For the following reaction, 2 AgNO 3(aq) + CaCl 2(aq) 2 AgCl (s) + Ca(NO 3 ) 2(aq) (a) What volume of M AgNO 3 is required to react completely with 250 ml of M CaCl 2? (b) What mass of AgCl should be produced? Part (a): volume of AgNO 3? 1st, find moles of CaCl 2 (250 ml) x mole 1,000 ml = mole CaCl 2 2nd, find moles of AgNO3 (0.100 mole CaCl 2 ) x 2 mole AgNO 3 1 mole CaCl 2 = mole AgNO 3 3rd, find volume of AgNO3 solution (0.200 mole AgNO 3 ) x 1,000 ml AgNO mole AgNO 3 = 800 ml AgNO 3 Page 6

7 Part (b): mass of AgCl? (0.100 mole CaCl 2 ) x 2 mole AgCl 1 mole CaCl 2 = mole AgCl (0.200 mole AgCl) x 143 g AgCl 1 mole AgCl = 28.6 g WORK MORE PROBLEMS!!! 5. Titrations Titration End-point Indicator An unknown amount of one reactant is combined exactly with a precisely measured volume of a standard solution of the other. When exactly stoichiometric amounts of two reactants have been combined. Substance added to aid in detection of the endpoint (usually via a color change) Vinegar is an aqueous solution of acetic acid, HC 2 H 3 O 2, which is often written as HAc for simplicity. A 12.5 ml sample of vinegar was titrated with a M solution of NaOH. The titration required ml of the base solution in order to reach the endpoint. What is the molar concentration of HAc in vinegar? NaOH (aq) + HAc (aq) NaAc (aq) + H 2 O (15.4 ml NaOH) x mole NaOH 1,000 ml x 1 mole HAc 1 mole NaOH M = = mole HAc mole HAc (12.5 ml) (1 L / 1000 ml) = M Page 7

8 Electrolytes 1. Dissociation Reactions of Salts (in aqueous solution) Electrolytes are solutes that produce ions in solution via dissociation (these solutions can conduct electricity) e.g., NaCl (s) Na + (aq) + Cl- (aq) (NH 4 ) 2 SO 4(s) 2 NH 4 + (aq) + SO 4 2- (aq) these are "strong" electrolytes % ionized some substances are "weak" electrolytes -- partially ionized (< 100%) or, "non-electrolytes" -- not ionized at all 2. Acids and Bases as Electrolytes Arrhenius acid-base concept Acid = H + supplier e.g., HNO 3, HCl, H 2 SO 4, etc. HNO 3(aq) H + (aq) + NO 3 - (aq) (see later section for "correct" reaction) Base = OH - supplier e.g., NaOH, Mg(OH) 2, etc. NaOH (s) Na + (aq) + OH- (aq) 3. Acid-Base Neutralization Reactions Acid + Base Salt + Water e.g., HNO 3(aq) + NaOH (aq) NaNO 3(aq) + H 2 O H 2 SO 4(aq) + 2 KOH (aq) K 2 SO 4(aq) + 2 H 2 O Page 8

9 4. Anhydrides (oxides) -- Not in the Textbook! Acidic Anhydrides -- nonmetal oxides hydrolyze to yield oxo acids! e.g., SO 3 + H 2 O H 2 SO 4 N 2 O 5 + H 2 O 2 HNO 3 Basic Anhydrides -- metal oxides hydrolyze to yield metal hydroxides! (i.e., bases) e.g., MgO (s) + H 2 O Mg(OH) 2(aq) K 2 O (s) + H 2 O 2 KOH (aq) 5. Ionization of Molecular Compounds Some molecular compds produce ions in solution via reactions with H 2 O e.g., HBr (g) + H 2 O H 3 O + (aq) + Br- (aq) HNO 3(aq) + H 2 O H 3 O + (aq) + NO 3 - (aq) 6. Strong and Weak Electrolytes Strong Electrolytes (100% ionized) Strong Bases: hydroxides of Group I or II metals e.g., NaOH, Ca(OH) 2, etc. Strong Acids: HCl hydrochloric acid [memorize] HBr hydrobromic acid HI hydroiodic acid HNO 3 nitric acid H 2 SO 4 sulfuric acid HClO 4 perchloric acid Page 9

10 Weak Electrolytes partially ionized via a "dynamic equilibrium" usually, the equilibrium state lies mainly on the reactant side Weak Acids: e.g., HF, HC 2 H 3 O 2, HNO 2, etc. HNO 2(aq) + H 2 O H 3 O + (aq) + NO 2 - (aq) Weak Bases: e.g., NH 3 NH 3(aq) + H 2 O NH 4 + (aq) + OH - (aq) Water itself is a weak electrolyte -- undergoes "autoionization" 2 H 2 O H 3 O + (aq) + OH- (aq) Ionic Reactions in Aqueous Solution 1. Equations for Ionic Reactions Metathesis Reaction (also called "double displacement") Ions from two different reactants simply trade partners, e.g.: Na 2 CO 3(aq) + Ba(NO 3 ) 2(aq) BaCO 3(s) + 2 NaNO 3(aq) This was written as a molecular equation in which all reactants and products are shown as complete, neutral chemical formulas. It could also have been written as a complete ionic equation in which all soluble ionic compounds are split up into their ions, e.g.: 2 Na + (aq) + CO 3 2- (aq) + Ba2+ (aq) + 2 NO 3 - (aq) BaCO 3(s) + 2 Na + (aq) + 2 NO 3 - (aq) Here, the Na + and NO 3 - ions are called "spectator ions" because they appear unchanged on both sides of the equation. Page 10

11 The spectator ions do not participate in the chemically important part of the reaction -- the precipitation of BaCO 3 The essential chemical process can be written without the spectator ions in the net ionic equation, e.g.: Ba 2+ (aq) + CO 3 2- (aq) BaCO 3(s) The net ionic equation shows that, in general, a precipitate of BaCO 3 will form whenever the ions Ba 2+ and CO 3 2- are combined in aqueous solution, regardless of their sources. 2. Summary of the three types of balanced chemical equations Molecular Equation shows all compounds with complete, neutral molecular formulas useful in planning experiments and stoichiometry calculations Ionic Equation (complete) all strong electrolytes are shown in their dissociated, ionic forms insoluble substances and weak electrolytes are shown in their molecular form "spectator ions" are included useful for showing all details of what is happening in the reaction Net Ionic Equation "spectator ions" are omitted only the essential chemical process is shown, i.e., formation of a: solid precipitate, gaseous product, or weak electrolyte (e.g., water) useful for generalizing the reaction -- same important product can often be formed from different sets of reactants When will a precipitate form?!!! KNOW THE SOLUBILITY RULES --- Table 4.1!!! Page 11

12 Oxidation-Reduction (Redox) Reactions 1. General Redox Concepts Redox reaction -- electron transfer process e.g., 2 Na + Cl 2 2 NaCl Overall process involves two Half Reactions: oxidation -- loss of electron(s) reduction -- gain of electron(s) e.g., Na Na + + e - (oxidation) Cl e - 2 Cl - (reduction) related terms: oxidizing agent = the substance that is reduced (Cl 2 ) reducing agent = the substance that is oxidized (Na) Oxidation and reduction always occur together so that there is no net loss or gain of electrons overall. 2. Oxidation Numbers (oxidation states) Oxidation Number: a charge that is assigned to an atom to aid in balancing redox reactions Generally, oxidation number is the charge that would result if all of the bonding electrons around an atom were assigned to the more electronegative element(s). Rules for assigning oxidation numbers -- see page 177 Learn the rules and practice many examples! Questions (1) Assign all oxidation numbers in: Ag 2 S ClO 3 - ClO 4 - Cr(NO 3 ) 3 H 2 O H 2 O 2 (2) Which of the following is a redox reaction? Determine what is being oxidized and reduced. Identify the oxidizing and reducing agents. 4 Al + 3 O 2 2 Al 2 O 3 CaO + CO 2 CaCO 3 Page 12

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