Chapter 2 Stoichiometry

1 8/2/2004 OFB Chapter 21 Chapter 2 Stoichiometry 2-1 Writing Balanced Chemical Equations 2-2 Using Balanced Chemical Equations 2-3 Limiting Reactant and Percentage Yield 2-4 The Stoichiometry of Reactions in Solution 2-5 the Scale of Chemical Processes8/2/2004 OFB Chapter 22 Chapter 2 Chapter 2--1 1 Balancing Chemical EquationsBalancing Chemical Equations Chemical Reactions tell us two things What atoms or molecules are reacting together to form other products How muchreactant & product are formed A Chemical reaction is a statement of experimental fact:KClO3 (s) KCl (s) + O2 (g) Reactants on left, products on right What is equation missing? Need balanced reaction Why?8/2/2004 OFB Chapter 23 Chemical Chemical EquationsEquations Because the same atoms are Because the same atoms are present in a reaction at the present in a reaction at the beginning and at the end, the beginning and at the end, the amount of matter in a system amount of matter in a system does not change.

2 Does not change. The The Law of the Conservation of Matter8/2/2004 OFB Chapter 24 Because of the principle of the Because of the principle of the conservation of matter, , It must have the same It must have the same number number of atoms of the same kind on of atoms of the same kind on both means an This means an equation must be balanced!Chemical Chemical EquationsEquations8/2/2004 OFB Chapter 25 Writing Balanced Chemical EquationsStep 1: Assign 1 as the coefficient of one reactant or product. The best choice is the most complicated species, with the largest number of different 2: Identify, in sequence, elements that appear in only one chemical species for which the coefficient is not yet determined. Choose that coefficient to balance the number of moles of atoms of that element.

3 Continue until all coefficients are 3: It is often desirable to eliminate fractional coefficients. To do so, multiply the entire equation by the smallest integer that eliminates the :8/2/2004 OFB Chapter 26 Balancing EquationsBalancing Equations8/2/2004 OFB Chapter 27 Limiting Reactants excesslimiting8/2/2004 OFB Chapter 28 Limiting ReactantShort-Cut Method1. Figure out the number of moles of every reactant. (moles = g per Molar Mass)2. Divide each answer by the coefficient that the reactant has in the balanced equation (moles per mole of reactant)3. The reactant for which the answer is the smallest is the limiting A + B A2B8/2/2004 OFB Chapter 29 Chemical Chemical EquationsEquations4 Al(s) + 3 O4 Al(s) + 3 O22(g) (g) ------> 2 Al> 2 Al22OO33(s)(s)This equation meansThis equation means4 Al atoms + 3 O4 Al atoms + 3 O22molecules molecules ------givegive------>>2 molecules of Al2 molecules of Al22OO33 OrOr4 moles of Al + 3 moles of O4 moles of Al + 3 moles of O22------givegive------>>2 moles of Al2 moles of Al22OO338/2/2004 OFB Chapter 210 Balancing Balancing EquationsEquations__C__C33HH88(g) +__ O(g) +__ O22(g) (g) __CO__CO22(g) + __H(g) + __H22O(g)O(g)

4 8/2/2004 OFB Chapter 211 Chapter 2 Chapter 2--1 1 Balancing Chemical Balancing Chemical EquationsEquations Tips for balancing equationsCoefficients2 H2O(l) 2 H2(g) + 1 O2(g)Subscripts Neverchange subscripts, only change molar coefficients. Balance simple equations by inspection Start with heaviest atom, balance, then next heaviest, etc. Balance H & O last, there is often H2O, H2, O2, OH , or H+in equations8/2/2004 OFB Chapter 212 Writing Balanced Chemical EquationsPbO2+ Pb + H2SO4 PbSO4+ H2 OProblem:Suppose we have grams of Pb in the presence of excess lead oxide and sulfuric acid. How many grams of Lead Sulfate are produced?8/2/2004 OFB Chapter 213 Mass of one reactant or product is knownMass of one reactant or product is unknownFinding the number of moles of the known speciesFinding the number of moles of the unknown speciesRatio of known to unknownRatio of known mass to its molar massRatio of unknown mass to its molar mass8/2/2004 OFB Chapter 214 Mass of one reactants or product is knownMass of one reactant or product is unknownFinding the number of moles of the known speciesFinding the number of moles of the un known speciesRatio of known to unknownRatio of known mass to its molar massRatio of unknown mass to its molar massPbO2+ Pb + 2H2SO4 2 PbSO4+ 2H2O8/2/2004 OFB Chapter 215 What mass (in grams) of the first reactant.

5 Would be required to react completely with g of the second reactant?XeF4+ 2 H2O Xe + 4 HF + g H2Ox g XeF4== g XeF41 mol g XeF41 mol H2O g H2O 1 mol XeF42 mol H2O Problem #16b, page 83:8/2/2004 OFB Chapter 216At one point in the purification of silicon, gaseous SiHCl3reacts with gaseous H2to give gaseous HCl and solid Si.(a) Determine the chemical amount (in moles) of H2required to react with mol of SiHCl3.(b) Determine the chemical amount of HCl that is produced.(c) Determine the mass (in grams) of Si that is (g) + H2(g) 3 HCl (g) + Si (s)8/2/2004 OFB Chapter 217At one point in the purification of silicon, gaseous SiHCl3reacts with gaseous H2to give gaseous HCl and solid Si.(a) Determine the chemical amount (in moles) of H2required to react with mol of SiHCl3.

6 (b) Determine the chemical amount of HCl that is produced.(c) Determine the mass (in grams) of Si that is (g) + H2(g) 3 HCl (g) + Si (s)8/2/2004 OFB Chapter 218 Volume Relationships of Gases in Chemical EquationsSome H2and N2 react to form L of NH3(g), according to the equation 3 H2(g) +N2(g) 2 N H3(g). What volume of H2(g) reacted, assuming that the pressure and temperature are the same after the reaction as before?Exercise 2-5:8/2/2004 OFB Chapter 219 The Stoichiometry of Reactions in SolutionThe actual amount of solute(the substance dissolved in the solvent) in any given volume of solution depends on how concentrated or dilute the solutionhappens to is expressed by a concentration(c) of a solute in a solution equals the chemical amount of the solute (n) divided by the volume (V) of the entire Chapter 220 Exercise 2-9 Calculate the molarityof a solution by dissolving g of Al(NO3)3in enough water to make mL of (NO3)3= g Al(NO3)3 (1 mol Al(NO3)3 / g Al(NO3)3)Vsolution= mL (1 L / 1000 mL) cAl(NO3)3= mol Al(NO3)3 / L Recall.

7 Molarity is expressed in moles per liter8/2/2004 OFB Chapter 221 Chapter 2 Stoichiometry Example / exercise all 2-1 to 2-12 Problems 2a-e, 4a-i, 14, 16a-d, 18, 20, 34, 38