Lecture 7

Lecture 7 - 4.4 Chemical Equa-ons Represent ...

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Unformatted text preview: 4.4 Chemical Equa-ons Represent Chemical Change Balancing Chemical Equa-ons Classifying Chemical Reac-ons Wri-ng Precipita-on Reac-ons as Net Ionic Equa-ons 4.5 Calcula-ons Using the Chemical Equa-on General Principles Use of Conversion Factors General, Organic and Biochemistry 7th Edition 4.4 Chemical Equa-ons Represent Chemical Change •  A chemical equa-on shows the molar quan-ty of reactants needed to produce a par-cular molar quan-ty of products. •  The whole- number coefficient before each formula in the equa-on gives rela-ve number of moles of each product and reactant in the reac-on. Ca C O •  Law of conserva0on of mass - maKer cannot be either gained or lost in the process of a chemical reac-on –  The total mass of the products must equal the total mass of the reactants Balancing Coefficient - how many of that substance are in the reac-on 2HgO( s ) ! → 2Hg(l ) + O 2 ( g ) ! Δ •  The equa-on is balanced if –  all the atoms of every reactant must also appear in the products Never H2Cl2 H2 + O2 → H2O WRONG: H2 + O2 → H2O2 Step 1. Count the number of moles of atoms of each element on both sides Reactants 2 mol H 2 mol O Products 2 mol H 1 mol O Step 2. Determine which elements are not balanced –Oxygen is not balanced Step 3. Balance one element at a -me by changing the coefficients H2 + O2 → 2H2O How will we balance hydrogen? 2H2 + O2 → 2H2O Step 4. Check! Make sure the law of conserva-on of mass is obeyed Reactants 4 mol H 2 mol O Products 4 mol H 2 mol O Practice Equation Balancing Balance the following equa-ons: C2H2 + O2 → CO2 + H2O 2C2H2 + 5O2 → 4CO2 + 2H2O AgNO3 + FeCl3 → Fe(NO3)3 + AgCl 3AgNO3 + FeCl3 → Fe(NO3)3 + 3AgCl C2H6 + O2 → CO2 + H2O 2C2H6 + 7O2 → 4CO2 + 6H2O N2 + H2 → NH3 N2 + 3H2 → 2NH3 Classifying Chemical Reac-ons Precipita-on Reac-ons •  Chemical change in a solu-on that results in one or more insoluble products •  To predict if a precipita-on reac-on can occur it is helpful to know the solubili-es of ionic compounds Predic-ng Whether Precipita-on Will Occur •  Recombine the ionic compounds to have them exchange partners •  Examine the new compounds formed and determine if any are insoluble according to the rules in Table 4.1 •  Any insoluble salt will be the precipitate Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2 ((?) + 2NaNO3 ( (?) ) s) aq Predict Whether These Reactions Form Precipitates •  Potassium chloride and silver nitrate KCl + AgNO3 à༎ AgCl (s) + KNO3 (aq) •  Potassium acetate and silver nitrate KOAc + AgNO3 à༎ KNO3 (aq) + AgOAc sodium hydroxide and ammonium chloride NaOH + NH4Cl à༎ sodium hydroxide and iron(II) chloride NaOH + FeCl2 à༎ Reac-ons with Oxygen •  Reac-ons with oxygen generally release energy •  Combus)on of natural gas –  Organic compounds CO2 and H2O are usually the products CH4+2O2→CO2+2H2O •  Rus)ng or corrosion of iron 4Fe + 3O2 → 2Fe2O3 Acid- Base Reac-ons •  These reac-ons involve the transfer of a hydrogen ion (H+) from one reactant (acid) to another (base) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) The H+ on HCl was transferred to the oxygen in OH- , giving H2O Oxida-on- Reduc-on Reac-ons •  Reac-on involves the transfer of one or more electrons from one reactant to another Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+(aq) Two electrons are transferred from Zn to Cu2+ Wri-ng Precipita-on Reac-ons as Net Ionic Equa-ons Wri-ng Net Ionic Equa-ons 4.5 Calcula-ons Using the Chemical Equa-on •  Calcula-on quan--es of reactants and products in a chemical reac-on has many applica-ons •  Need a balanced chemical equa-on for the reac-on of interest •  The coefficients represent the number of moles of each substance in the equa-on General Principles 1.  2.  Chemical formulas of all reactants and products must be known Equa-on must be balanced to obey the law of conserva-on of mass •  Calcula-ons of an unbalanced equa-on are meaningless 3.  Calcula-ons are performed in terms of moles •  Coefficients in the balanced equa-on represent the rela-ve number of moles of products and reactants Using the Chemical Equa-on •  Examine the reac-on: 2H2 + O2 → 2H2O •  Coefficients tell us? 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O •  What if 4 moles of H2 reacts with 2 moles of O2? It yields 4 moles of H2O •  The coefficients of the balanced equa-on are used to convert between moles of substances •  How many moles of O2 are needed to react with 4.26 moles of H2? •  Use the factor- label method to perform this calcula-on _1_mol O 2 4.26 mol H 2 × = 2 __ mol H 2 2.13 mol O2 •  Digits in the conversion factor come from the balanced equa-on Conversion Between Moles and Grams •  Requires only the formula weight •  Convert 1.00 mol O2 to grams –  Find the molar mass of oxygen •  32.0 g O2 = 1 mol O2 –  Set up the equa-on –  Cancel units Solve equa-on 1.00 × 32.0 g O2 = 32.0 g O2 Conversion of Mole Reactants to Mole Products •  Use a balanced equa-on •  C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) •  1 mol C3H8 results in: –  5 mol O2 consumed 1 mol C3H8 /5 mol O2 –  3 mol CO2 formed 1 mol C3H8 /3 mol CO2 –  4 mol H2O formed 1 mol C3H8 /4 mol H2O •  This can be rewriKen as conversion factors Calculate grams O2 reacting with 1.00 mol C3H8 Use 2 conversion factors –  Moles C3H8 to moles O2 moles moles –  Moles of O2 to grams O2 C3H8 Oxygen grams Oxygen Calcula-ng Grams of Product from Moles of Reactant •  Calculate grams CO2 from combus-on of 1.00 mol C3H8 •  Use 2 conversion factors moles moles –  Moles C3H8 to moles CO2 C3H8 CO2 –  Moles of CO2 to grams CO2 Rela-ng Masses of Reactants and Products grams H2O moles H2O moles C3H8 grams C3H8 •  Calculate grams C3H8 required to produce 36.0 grams of H2O •  Use 3 conversion factors –  Grams H2O to moles H2O –  Moles H2O to moles C3H8 –  Moles of C3H8 to grams C3H8 grams CO2 Calcula-ng a Quan-ty of Reactant •  Ca(OH)2 neutralizes HCl •  Calculate grams HCl neutralized by 0.500 mol Ca(OH)2 –  Write chemical equa-on and balance •  Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l) moles moles grams Ca(OH)2 HCl HCl General Problem- solving Strategy What mass of sodium hydroxide, NaOH, would be required to produce 8.00 g of the antacid milk of magnesia, Mg(OH)2, by the reac-on of MgCl2 with NaOH? X g 2 mols 8 g 1 mol (X/40.0 ): (8/58.3) = 2:1 X/40.0 = 2x(8/58.3) X = 10.97777 =11.0 g NaOH: 40.0 g/mol Mg(OH)2: 58.3 g/mol Sample Calcula-on Na + Cl2 → NaCl 1.  Balance the equa-on 2Na + Cl2 → 2NaCl 2.  Calculate the moles Cl2 reac-ng with 5.00 mol Na 3.  Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2 4.  Calculate the grams Na reac-ng with 5.00 g Cl2 Theore-cal and Percent Yield •  Theore-cal yield - the maximum amount of product that can be produced –  Pencil and paper yield •  Actual yield - the amount produced when the reac-on is performed –  Laboratory yield •  Percent yield: 132 g of CO2 produced from the combus-on of 1.00 mol of C3H8. actual yield % yield = ×100% = 125 g CO2 actual × 100% = 97.4% theoretical yield 132 g CO2 theore-cal If the theore-cal yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield: 2 Al(s) + Fe2O3(s) → Al2O3(aq) + 2Fe(aq) •  [25.0 g / 30.0 g] × 100% = 83.3% •  Calculate the % yield if 26.8 grams iron was collected in the same reac-on How many g of NO2 can be produced by the reac-on of excess O2 with 50.0 g of NO? If the actual yield of NO2 is 50.0 g, what is the % yield? ...
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This note was uploaded on 03/07/2012 for the course 830 201 taught by Professor Leyton during the Fall '08 term at Rutgers.

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