• Textbook Reading: 

    A mole working at a lab bench

    Holt "Chemistry" Chapter 9 Sections 1-2, Chapter 13 Section 2
    Silberberg "Chemistry" Chapter 3 Sections 3-4, Chapter 4 Section 1

     Balanced Chemical Equation Is

    • A chemical equation which has equal numbers of each kind of atoms on both sides.
      • This is a consequence of conservation of mass.
      • Unbalanced Equation Example: C2H5OH + I2 + Na2CO3 + H2O → CHI3 + NaHCO2 + CO2 + HI
        • Reactants:  3 C atoms, 2 I atoms, 2 Na atoms, 5 O atoms, 7 H atoms
        • Products:  3 C atoms, 4 I atoms, 1 Na atom, 4 O atoms, 3 H atoms
      • Balanced Equation Example: 2 C2H5OH + 8 I2 + Na2CO3 + H2O → 2 CHI3 + 2 NaHCO2 + CO2 + 10 HI
        • Both Sides (Reactants & Products):  5 C atoms, 16 I atoms, 2 Na atoms, 6 O atoms, 14 H atoms
    • An equation is balanced by inserting coefficients before formulas as necessary.
      • If there is no coefficient, a '1' is implied.
      • Coefficients are generally integers (although you may see some exceptions when we deal with thermochemistry).
      • Coefficients should not all share a common factor:
        • 6 H2 + 2 N2 → 4 NH3 : BAD because all coefficients are multiples of 2. 
        • 3 H2 + N2 → 2 NH3 : GOOD because 3, 1, 2 have no common factor.
    • Balancing is an art but there are helpful rules.
      • "Double the odds":  if an element is present in multiple formulas in even numbers everywhere but one place where it is odd, double the coefficient for the formula where it is odd.
        • H2 + O2 → H2O:  "Double the odd" oxygen atom in H2O to get H2 + O2 → 2 H2O. 
        • Now balance H2 to get the balanced equation 2 H2 + O2 → 2 H2O
      • "Least common multiple":  if an element is present on both sides in 2 different numbers, bring them both up to the LCM.
        • KClO3 → KCl + O2:  "LCM" of 2 and 3 is 6 oxygens.  
        • 2 KClO3 → KCl + 3 O2
        • Now balance Cl and K:2 KClO3 → 2 KCl + 3 O2
        • H2 + O2 → H2O : do not modify H2O to get H2 + O2 → H2O2
        • This is no longer talking about the same substances, so it is wrong.
    • Practice your balancing.  Play The Game!  This is a regular class activity, but you can get ahead if you like.


    Coefficients of a Balanced Chemical Equation are Ratios of Moles

    • This sounds familiar.  Didn't we say that Chemical Formulas are Ratios of Moles?
    • The equation 3 H2 + N2 → 2 NH3 means 3 moles of hydrogen react with 1 mole of nitrogen producing 2 moles of ammonia.
    • We can write the mole ratios (3 mol H2/1 mol N2), (3 mol H2/2 mol NH3), and (1 mol N2/2 mol NH3) for example.
    • We can use these ratios to relate between any products and reactants in the same reaction.
      • If you have 9 moles of H2, you will require 3 moles of N2 to react with it.
      • You will obtain 6 moles of NH3 if this reaction goes perfectly.
    • The relationships between quantities of products and reactants are known as stoichiometry.


    Theoretical and Percentage Yield

    • The theoretical yield of a reaction is the amount of a product predicted from the stoichiometry of the balanced equation.
    • The actual yield of a reaction is the amount of product you get in a real experiment.
    • The percentage yield of a reaction is 100 x actual yield / theoretical yield.
      • The equation 3 H2 + N2 → 2 NH3 predicts that 18 moles of H2 and 6 moles of N2 yield 12 moles of NH3.
      • If you actually do the reaction but only obtain 9 moles of NH3, the theoretical yield is 100 x 9 / 12 = 75%.


    Limiting and Excess Reactants

    • The grocery store sells hotdogs in packs of 10, but buns only come in packs of 8.
    • If you buy one of each, you can only have 8 sandwiches.
    • We would say that buns are limiting but hotdogs are in excess, because there will be some hotdogs left over.
    • The yield of a reaction is determined by the limiting reactant.  
    • Suppose we have 6 moles of H2 and 4 moles of N2 and we react them as in 3 H2 + N2 → 2 NH3
      • 6 moles of H2 is enough to produce 4 moles of NH3.
      • 4 moles of N2 is enough to produce 8 moles of NH3 but we won't get that: instead, some N2 will be left over.
      • We get the smaller yield, 4 moles of NH3.  H2 is the limiting reactant and N2 is in excess.


    Stoichiometry in Solution

    • Chemical reactions are often conducted in solution.  
    • The molarity  of a solution is defined as the number of moles of a substance per Liter of solution.
      • M = n/V
    • The symbol for molarity is a capital M.
      • 3M HCl means a solution containing 3 moles of HCl per Liter
      • You can write the ratio (3 mol HCl/1 L) for use in stoichiometry calculations.


     Use Mole Ratios from the Balanced Chemical Equation to Solve Stoichiometry Problems


    Iron (III) hydroxide decomposes when heated to yield iron (III) oxide and water vapor by the following equation: 

                  2 Fe(OH)3 → Fe2O3 + 3 H2O

    How many moles of water vapor are released by the decomposition of 8 moles of Fe(OH)3?

     8 mol Fe(OH)3  x 3 mol H2O      = 12 mol H2O
                      2 mol Fe(OH)3


    Aluminum reacts with oxygen by the equation 4 Al + 3 O2 → 2 Al2O3.  What mass of oxygen (O2) is required to react with 18 g of aluminum?

    18 g Al x  1 mol Al  x 3 mol O2 x 32.0 g O2 = 16 g O2
              27.0 g Al    4 mol Al    1 mol O2

    Iron (Fe) dissolves in dilute sulfuric acid (H2SO4) yielding iron (II) sulfate and hydrogen gas.  How much 3 M H2SO4 (3 moles per Liter) solution is required to react with 100 g of Fe metal?

    Write a balanced chemical equation:  Fe + H2SO4 → FeSO4 + H2

     100 g Fe  x 1 mol Fe    x 1 mol H2SO4 x 1 L H2SO4 solution = 0.597 L H2SO4 solution
                 55.85 g Fe    1 mol Fe      3 mol H2SO4

    Find Limiting Reactant by Working Out Stoichiometry for each Reactant 

    How many grams of nitric acid, HNO3, can be prepared from the reaction of 69.0 g of NO2 with 29.0 g H2O according to the equation below?  What is the limiting reactant?  Which reactant is in excess?

    3 NO2 + H2O → 2HNO3 + NO

    Work out the yield of HNO3 starting from NO2

    69.0 g NO2 x 1 mol NO2  x 2 mol HNO3 x 63.0 g HNO3 = 63.0 g HNO3
    46.0 g NO2 3 mol NO2 1 mol HNO3

    Work out the yield of HNO3 starting from H2O

    29.0 g H2O x 1 mol H2O  x 2 mol HNO3 x 63.0 g HNO3 = 203 g HNO3
    18.0 g NO2 1 mol H2O 1 mol HNO3

    The theoretical yield is the smaller of the two, 63.0 g HNO3, from NO2.  The limiting reactant is NO2 because it limits the amount of product.  Not all the H2O will react, so H2O is in excess.

    Use Reaction Stoichiometry to Calculate Theoretical Yield

    Suppose the reaction above between NO2 and H2O  was performed, but instead of the theoretical yield of 63.0 g HNO3 , only 50.0 g of HNO3 was obtained.  No real reaction is perfect.  What is the percentage yield of this reaction?

    100 x 50.0 g HNO3 = 79.3 %
          63.0 g HNO3

    Solve Stoichiometry Problems for Unknown Molar Masses

    Remember that the molar mass of a substance is equal to the mass in grams of a sample of that substance, divided by the number of moles of that substance.

    22.4 grams of an unknown metal carbonate, represented as MCO3, is heated to decompose it to the metal oxide MO and carbon dioxide.  8.8 grams of carbon dioxide are produced by the reaction MCO3 → MO + CO2.  What is the identity of the metal M?

    8.8 g CO2 x  1 mol CO2  x  1 mol MCO3  = 0.2 mol MCO3
    44 g CO2 1 mol CO2

    22.4 g MCO3  = 112 g/mol : this is the molar mass of MCO3
    0.2 mol MCO3

    Subtracting the mass of CO3 = (12.0 + 3 * 16.0) = 60 g/mol, the molar mass of M must be 112 - 60 = 52 g/mol.  The metal M must therefore be chromium (Cr).

    Take the Test!