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Reaction Stoichiometry: Mole Method Calculations

Reaction Stoichiometry: Mole Method Calculations. Ch. 4: Chemical Quantities and Aqueous Reactions. Coefficients in balanced equations give the ratio by moles!!! 2 C 4 H 10 + 13 O 2 8 CO 2 + 10 H 2 O e.g. in the above reaction;

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Reaction Stoichiometry: Mole Method Calculations

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  1. Reaction Stoichiometry: Mole Method Calculations Ch. 4: Chemical Quantities and Aqueous Reactions Coefficients in balanced equations give the ratio by moles!!! 2 C4H10 + 13 O2 8 CO2 + 10 H2O e.g. in the above reaction; 2 moles C4H10 react with 13 moles O2 to produce 8 moles CO2 and 10 moles H2O Use these just like other conversion factors! Problem How many moles of O2 are required to react with 0.500 moles of C4H10 according to the balanced equation? 0.500 mol C4H10 x = 3.25 mol O2 13 mol O2 2 mol C4H10

  2. Example Problem What mass of CO2 could be produced from the combustion of 100 g (3 sig fig) of butane (C4H10)? Always convert given quantities (e.g. grams) to moles!!! grams A moles A moles B grams B will need formula masses to convert between grams and moles: CO2 = 44.01 g/mol C4H10 = 58.12 g/mol Step 1: “grams A moles A” 100 g C4H10 x = 1.721 mol C4H10 Step 2: “moles A moles B” 1.721 mol C4H10 x = 6.884 mol CO2 1 mol C4H10 58.12 g C4H10 8 mol CO2 2 mol C4H10 continued on next slide…

  3. Example Problem, cont. Step 3: “moles B grams B” 6.884 mol CO2 x = 3.03 x 102 g CO2 Alternatively, the 3 steps can be combined into one unit analysis equation: 100 g C4H10 x x x = 3.03 x 102 g CO2 44.01 g CO2 1 mol CO2 1 mol C4H10 8 mol CO2 44.01 g CO2 58.12 g C4H10 2 mol C4H10 1 mol CO2

  4. Limiting Reactant Calculations • In practice, reagents 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!!!

  5. Example Problem In a commercial process, nitric oxide (NO) is produced: 4 NH3 + 5 O2 4 NO + 6 H2O What mass (in grams) of NO can be made from the reaction of 30.00 g NH3 and 40.00 g O2? 1st, find moles of reactants: 30.00 g NH3 x = 1.762 mol NH3 40.00 g O2 x = 1.250 mol O2 2nd, calculate amount of product based on each reactant, separately: Yield of NO based on NH3: 1.762 mol NH3 x = 1.762 mol NO 1 mol NH3 17.03 g NH3 1 mol O2 32.00 g O2 4 mol NO 4 mol NH3 continued on next slide…

  6. Example Problem, cont. Yield of NO based on O2: 1.250 mol O2 x = 1.000 mol NO Therefore, O2 is the limiting reactant! (excess of NH3 exists) A maximum of 1.000 moles of NO can be produced. 3rd, calculate yield of product based on limiting reactant: 1.000 mol NO x = 30.01 g NO 4 mol NO 5 mol O2 30.01 g NO 1 mol NO

  7. 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%

  8. Example Problem In the previous experiment for the production of NO from 30.00 g NH3 and 40.00 g O2, the chemist obtained 25.50 g NO. What is the percentage yield of this reaction? 4 NH3 + 5 O2 4 NO + 6 H2O Theoretical yield = 30.01 g NO (based on limiting reactant) Actual yield = 25.50 g (given in problem) % yield = (25.50 g NO/30.01 g NO) x 100% = 85.0% (Note: the % yield can never be more than 100%!)

  9. Sample Problem The simplest borane known as “diborane-6,” B2H6, can be prepared from boron trichloride and lithium hydride according to the following balanced equation. 2 BCl3 + 6 LiH B2H6 + 6 LiCl In one experiment, 20.0 g of BCl3 was allowed to react with 5.00 g of LiH. After isolation and purification of the reaction product, 2.05 g of pure B2H6 was actually obtained. Calculate the percentage yield of this reaction. Formula masses (g/mol): BCl3 = 117.2 LiH = 7.95 B2H6 = 27.7 LiCl = 42.4

  10. Reactions in Aqueous Solution Solution Terminology solution -- homogeneous (uniform) mixture consisting of: solvent - the bulk medium, e.g. H2O, solute(s) - the dissolved substance(s), e.g. NaCl concentration -- measure of relative solute/solvent ratio --concentrated vs. dilute 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 H2O (“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) + AgNO3(aq) --> AgCl(s) + NaNO3(aq)

  11. Solution Concentration • 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 Problem What mass of NaCl is required to prepare 300 mL of 0.100 M solution? 1st--find moles of NaCl required: 300 mL x = 0.0300 mol NaCl 2nd--convert to grams of NaCl: 0.0300 mol NaCl x = 1.75 g NaCl Prepare this solution by weighing 1.75 g NaCl, dissolving in some H2O (about 250 mL), and then diluting to the 300 mL mark. 0.100 mol NaCl 1000 mL solution 58.44 g NaCl 1 mole NaCl

  12. Sample Problems Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate in enough water to form 125 mL of solution. The label of a stock bottle of aqueous ammonia indicates that the solution is 28% ammonia by mass and has a density of 0.898 g/mL. Calculate the molarity of the solution.

  13. Dilution of Concentrated Solutions Concentrated solution + H2O --> dilute solution (moles solute)conc = (moles solute)dil VcMc = VdMd Problem A 5.00 M NaCl “stock” solution is available. How would you prepare 300 mL of a 0.100 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 H2O to a total volume of 300 mL.

  14. Stoichiometry Problems -- Reactions in Solution Start with a Balanced Equation and Use the Mole Method (Molarity is just a conversion factor!) Problem For the following reaction, 2 AgNO3(aq) + CaCl2(aq) --> 2 AgCl(s) + Ca(NO3)2(aq) a) What volume of 0.250 M AgNO3 is required to react completely with 250 mL of 0.400 M CaCl2? b) What mass of AgCl should be produced? Part (a): volume of AgNO3? 1st, find moles of CaCl2 (250 mL CaCl2 soln) x = 0.100 mol CaCl2 2nd, find moles of AgNO3 (0.100 mol CaCl2) x = 0.200 mol AgNO3 0.400 mol CaCl2 1000 mL CaCl2 soln 2 mol AgNO3 1 mol CaCl2

  15. Example Problem, cont. 3rd, find volume of AgNO3 solution (0.200 mole AgNO3) x = 800 mL AgNO3 solution Part (b): mass of AgCl? 0.100 mol CaCl2 x = 0.200 mol AgCl (0.200 mol AgCl) x = 28.6 g AgCl WORK MORE PROBLEMS!! 1000 mL AgNO3 soln 0.250 mol AgNO3 2 mol AgCl 1 mol CaCl2 143 g AgCl 1 mol AgCl

  16. Formation of a Solution No Dissociation; I2 in H2O Dissociation; NaCl in H2O Formation of a solution of iodine molecules in ethyl alcohol. Ethyl alcohol is the solvent and iodine the solute. Na+ and Cl-ions are stabilized when they are surrounded by H2O molecules. • Water is probably the most common solvent. When a substance is • dissolved into water it is said to be aqueous. (from the Latin, aqua, water).

  17. Electrolytes Dissociation Reactions of Salts (in aqueous solution) Electrolytes are solutes that produce ions in solution via dissociation (their solutions can conduct electricity), e.g. • NaCl(s) --> Na+(aq) + Cl–(aq) • (NH4)2SO4(s) --> 2 NH4+(aq) + SO42–(aq) These are “strong” electrolytes -- 100% ionized (e.g. “soluble” salts, strong acids) Some substances are “weak” electrolytes -- partially ionized (< 100%) (e.g. weak acids) or, “non-electrolytes” -- not ionized at all (e.g. sugar, iodine)

  18. Solubility Rules for Ionic Compounds in Water I • Soluble Compounds

  19. Solubility Rules for Ionic Compounds in Water II • Insoluble Compounds A knowledge of these rules will allow you to predict a large number of precipitation reactions

  20. Sample Questions Write the equations for the dissociation of the following ionic compounds. a) KBr b) Na2SO4 c) ZnCl2

  21. Precipitation Reactions When two dissociated ionic substances combine, an insoluble precipitate may form. e.g. the reaction between lead(II) nitrate and potassium iodide, below, forms insoluble lead(II) iodide. +

  22. Ionic Reactions in Aqueous Solution • Equations for Ionic Reactions Metathesis Reaction (also called “double displacement”) Ions from two different reactants simply trade partners, e.g.: Na2CO3(aq) + Ba(NO3)2(aq) --> BaCO3(s) + 2 NaNO3(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) + CO32–(aq) + Ba2+(aq) + 2 NO3–(aq) --> BaCO3(s) + 2 Na+(aq) + 2 NO3–(aq)

  23. Equations for Ionic Rxns, cont. 2 Na+(aq) + CO32–(aq) + Ba2+(aq) + 2 NO3–(aq) --> BaCO3(s) + 2 Na+(aq) + 2 NO3–(aq) Here, the Na+ and NO3– ions are called “spectator ions” because they appear unchanged on both sides of the equation. The spectator ions do not participate in the chemically important part of the reaction -- the precipitation of BaCO3. The essential chemical process can be written without the spectator ions in the net ionic equation, e.g.: Ba2+(aq) + CO32–(aq) --> BaCO3(s) The net ionic equation shows that, in general, a precipitate of BaCO3 will form whenever the ions Ba2+ and CO32– are combined in aqueous solution, regardless of their sources. Both atoms and charge must balance!

  24. Strategy • Dissociate the ionic compounds into their ions. • Look at the reaction and see what the combinations of ions could be. • Consult your memory or solubility charts to determine the solubility of the product. • Balance the equation(s). Example Problem Will a precipitate form when aqueous solutions of CaCl2 and K3PO4 are mixed? Write balanced molecular, ionic, and net ionic equations for the reaction. Step 1: Write the possible ions: Ca2+, Cl–, K+, PO43– Step 2: Look at the combinations; KCl, Ca3(PO4)2 Step 3: You know KCl is soluble, Ca3(PO4)2 is not, so a precipitate will form. Step 4: Balance the equation: 3 CaCl2 + 2 K3PO4 --> Ca3(PO4)2 + 6 KCl

  25. Example Problem, cont. Molecular equation is: 3 CaCl2(aq) + 2 K3PO4(aq) --> Ca3(PO4)2(s) + 6 KCl(aq) The ionic equation: 3 Ca2+(aq) + 6 Cl–(aq) + 6 K+(aq) + 2 PO43–(aq) --> Ca3(PO4)2(s) + 6 K+(aq) + 6 Cl–(aq) Ions that do not participate in the reaction are called spectator ions. Net ionic equation is: 3 Ca2+(aq) + 2 PO43–(aq) --> Ca3(PO4)2(s)

  26. Summary of 3 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

  27. Summary of 3 Types of Balanced Chemical Equations, cont. • 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

  28. Sample Questions Write balanced molecular equations for the following reactions. a) AgNO3 + NaCl --> b) BaCl2 + Na2SO4 --> c) Pb(NO3)2 + NaC2H3O2 --> d) NaOH + nickel(II) nitrate --> e) AgNO3 + Na2S -->

  29. Arrhenius Acid-base Concept Acid = H+ supplier e.g. HNO3, HCl, H2SO4, etc In water, HNO3 forms H+(aq) + NO3–(aq) Base = OH– supplier e.g. NaOH, Mg(OH)2, etc. In water, NaOH forms Na+(aq) + OH–(aq) • Acid-Base Neutralization Reactions Acid + Base --> Salt + Water e.g. HNO3(aq) + NaOH(aq) --> NaNO3(aq) + H2O(l) H2SO4(aq) + 2 KOH(aq) --> K2SO4(aq) + 2 H2O(l)

  30. Sample Problem • Write a balanced chemical equation for the complete neutralization reaction of barium hydroxide with sulfuric acid. Circle the species in your equation that is the “salt.”

  31. Titrations Titration An unknown amount of one reactant is combined exactly with a precisely measured volume of a standard solution of the other. End-point When exactly stoichiometric amounts of two reactants have been combined. Indicator Substance added to aid in detection of the endpoint (usually via a color change)

  32. Example Problem Vinegar is an aqueous solution of acetic acid, HC2H3O2. A 12.5 mL sample of vinegar was titrated with a 0.504 M solution of NaOH. The titration required 15.40 mL of the base solution in order to reach the endpoint. What is the molar concentration of HC2H3O2 in vinegar? NaOH(aq) + HC2H3O2(aq) --> NaC2H3O2(aq) + H2O(l) (15.40 mL NaOH) x x = 0.007762 mol HC2H3O2 M = = 0.621 M acetic acid 0.504 mol NaOH 1 mole HC2H3O2 1 mole NaOH 1000 mL NaOH soln 0.007762 mol HC2H3O2 0.0125 L vinegar

  33. Sample Problem A new fluorocarbon material recently prepared at TCU is believed to be a diprotic acid (call it H2A for short). A 0.860 g sample of this unknown acid was titrated (reaction below) with a 0.1025 M solution of NaOH. In the titration, 38.50 mL of the NaOH solution was required to reach the endpoint. Calcuate the molecular mass of the new acid. H2A + 2 NaOH --> Na2A + 2 H2O

  34. Gas Evolution Reactions

  35. Sample Questions Write the molecular and net ionic equations for the reactions below. a) aluminum hydroxide + hydrochloric acid b) calcium carbonate + hydrochloric acid

  36. Oxidation - Reduction Reactions Redox Terminology redox reaction -- electron transfer process e.g., 2 Na + Cl2 --> 2 NaCl Related terms: oxidizing agent = the substance that gets reduced (Cl2) reducing agent = the substance that gets oxidized (Na) Oxidation and reduction always occur together so that there is no net loss or gain of electrons overall.

  37. OILRIG • O Oxidation • I Is • L Loss • R Reduction • I Is • G Gain

  38. Oxidation Numbers (Oxidation States) Oxidation number a “charge” that is assigned to an atom to aid in balancing redox reactions The oxidation state is the atomic number (no. of positive charges) minus the number of electrons that it is said to have. Oxidation: the oxidation state increases. Reduction: the oxidation state decreases. Rules for assigning oxidation numbers -- see page 164 Learn the rules by working examples!

  39. Assigning Oxidation States • Oxidation state of a free element is zero. • Oxidation state of a simple monoatomic ion is equal to the charge on the ion. • Sum of all the oxidation numbers must equal the charge on the particle. • In its compounds, H and Group 1A metals have an oxidation number of +1. Group 2A metals have an oxidation number of +2. • In its compounds, fluorine has an oxidation state of -1. • In its compounds, oxygen has an oxidation state of -2. If there is a conflict between two rules, apply the rule with the lower number and ignore the conflicting rule. e.g. assign all oxidation numbers in: Ag2S ClO3– ClO4– Cr(NO3)3 H2O H2O2

  40. Assigning Oxidation States • Oxidation state of a free element is zero. • Oxidation state of a simple monoatomic ion is equal to the charge on the ion. • Sum of all the oxidation numbers must equal the charge on the particle. • In its compounds, H and Group 1A metals have an oxidation number of +1. Group 2A metals have an oxidation number of +2. • In its compounds, fluorine has an oxidation state of -1. • In its compounds, oxygen has an oxidation state of -2. If there is a conflict between two rules, apply the rule with the lower number and ignore the conflicting rule. e.g. assign all oxidation numbers in: Ag2S ClO3– ClO4– Cr(NO3)3 H2O H2O2 -2 +5 -2 +1 +7 -2 +3 +5 -2 +1 -2 +1 -1

  41. Sample Problems --Note that fractional values of oxidation numbers are allowed. More examples: NH3 PCl5 FeS Na2O Mg2TiO4 V(C2H3O2)3 HPO42– K2Cr2O7

  42. Redox Reactions • The transfer of electrons between species, meaning that one species is oxidized and one is reduced. The two processes will always occur together. • When has a redox reaction occurred? • If there is a change in the oxidation state of any element in the reaction, a redox reaction has happened. • Remember that if something is oxidized, something else must be reduced! (and vice-versa) More examples: 2 Ca(s) + O2(g) --> 2 CaO(s) 2 Al(s) + 3 Cl2(g) --> 2 AlCl3(s)

  43. Example Problems • When H2O2 (hydrogen peroxide) is used as an antiseptic it oxidizes bacteria. What product might it form, O2 or H2O? • Answer: In the starting material O is in the -1 oxidation state. If it oxidizes bacteria, it must get reduced, which means its oxidation number goes down. O2 has an oxidation number of 0, while the O in H2O has an oxidation number of -2. Therefore H2O is the product of reduction of hydrogen peroxide. In the following reactions, state which compounds are oxidized and which are reduced. • 2 Al(s) + 3 I2(s) --> 2 AlI3(s) • 2 PbS(s) + 3 O2(g) --> 2 PbO(s) + 2 SO2(g) • 2 KCl(aq) + MnO2(s) + 2 H2SO4(aq) --> K2SO4(aq) + MnSO4(aq) + Cl2(g) + 2 H2O(l) • Answer: Oxidized: Al, S, Cl, and Reduced: I, O, Mn.

  44. Sample Problem In the following redox reaction, 14 HNO3(aq) + 4 K2C2O4(aq) + K2Cr2O7(aq) --> 8 CO2(g) + 2 Cr(NO3)2(aq) + 10 KNO3(aq) + 7 H2O(l) A) what is the substance oxidized? B) what is the oxidizing agent? C) what are the oxidation numbers of each atom in Cr(NO3)2?

  45. Combustion Reactions ¤ Combustion of Hydrocarbons (produces CO2 and H2O) e.g., 2 C8H18(l) + 25 O2(g) --> 16 CO2(g) + 18 H2O(l) 2 C2H5OH(l) + 6 O2(g) --> 4 CO2(g) + 6 H2O(l) ¤ Other Combustion Reactions e.g., N2(g) + O2(g) 2 NO(g) 4 Na(s) + O2(g)  2 Na2O(s)

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