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Typical Approach to Stoichiometry
 Very algorithmic
   grams A --> moles A--> moles B--> grams B
   Based on factor-label, unit cancelling, dimensional analysis
   Fosters “plug-n-chug” solution
   Approach can be used without much conceptual understanding
      Disconnected from balanced equation and physical reaction
      Relies exclusively on computation ability and favors math-strong students
      Especially poor for limiting reactant problems
BCA Approach
 Stresses mole relationships based on coefficients in
  balanced chemical equation
 Allows students to reason through stoichiometry

 Sets up students for equilibrium calculations later (ICE
  tables)
 Clearly shows limiting reactants
Emphasis on balanced equation
 Step 1- Balance the equation
  Hydrogen sulfide gas, which smells like rotten eggs, burns in air to produce sulfur
  dioxide and water.

 How many moles of oxygen gas would be needed to
  completely burn 2.4 moles of hydrogen sulfide?
                 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
       Before
       Change
       After
Focus on mole relationships

 Step 2: Fill in the “Before” line
   Assume more than enough O2 to react




             2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
   Before        2.4           XS         0   0
   Change
   After
Focus on mole relationships
 Step 3: Use ratio of coefficients to determine change
   This is done using “for every” statement proportional reasoning statements
       E.g., according to the reaction, “for every 2mol of hydrogen sulfide, 3mol of oxygen is
        required”


 Reactants are consumed (-) and products are formed (+)

              2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
   Before         2.4             XS                0            0
   Change         -2.4            -3.6           +2.4          +2.4
   After
Emphasize that change and after
     are not equivalent
 Step 4: Complete the table




           2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
  Before     2.4      XS        0      0
  Change    -2.4     -3.6      +2.4   +2.4
  After       0       XS       2.4    2.4
Complete other calculations on the
                side
 In this case, desired answer is in moles

 If mass is required, calculate from moles to grams in your
  usual way




                                 32 .0 γ
                 3.6 µολεσ 2 ×
                          Ο              =115 γ
                                 1 µολε
Only moles go in the BCA table
 The balanced equation deals with:

               how many, not how much
 If given mass of reactants:
   Calculate to moles first, then use the table

 If asked for mass of products:
   Use table first, then calculate to grams
Limiting Reactant Problems
 BCA approach distinguishes between what you start with and
  what reacts
  When 0.50 mol of aluminum reacts with 0.72 mol of iodine to form aluminum iodide,
  -How many moles of the excess reactant will remain?
  -How many moles of aluminum iodide will be formed?



                   2 Al +     3 I2 ---->   2 AlI3
        Before       0.5        0.72          0
        Change
        After
Limiting Reactant Problems
 Assume first reactant is all used, then reason how many moles
  of the other reactant you would need and compare to what you
  have:
     For every 2 mol of aluminum, you need 3 mol of iodine; for 0.5mol of aluminum, you’d need 0.75 mol
      of iodine
       Since you only have 0.72 mol of iodine, you “don’t have enough” and the iodine will limit the
           products
            Therefore, iodine is the limiting reactant and is totally used up in the reaction


                       2 Al +        3 I2 ---->       2 AlI3
         Before          0.5           0.72              0
         Change          -0.5          -0.75
         After
 It is thus clear to students that there’s not enough I2 to react
  with all the Al
Limiting Reactant Problems
 Now, proceed with BCA table calculation based on iodine being
  totally used, and determine the desired quantity(ies):
   0.02mol of aluminum are left (could calculate grams if needed)
   0.48mol of aluminum iodide are formed (could calculate grams)


                2 Al +   3 I2 ---->   2 AlI3
       Before    0.5       0.72         0
       Change   -0.48     -0.72       +0.48
       After     0.02        0        0.48
BCA Table = Versatile Tool
 It doesn’t matter what are the units of the initial given
  quantities in a stoichiometry problem:
   Mass - use molar mass
   Gas volume - use molar volume
                                                       1 µολε
                                          1.6 γ Ο2 ×           = 0.050 µολεσ
   Solution volume - use molarity                     32 .0 γ
                                                       1 µολε
                                           7.84 Λ×             = 0.350 µολεσ
                                                       22 .4 Λ
                                                       0.100 µολε
                                         0.0250 Λ×                = 0.00250 µολε
                                                          1.0 Λ




 Calculate to find moles, then use the BCA table

 Solve for how many first, and then for how much
BCA Table Template
Balanced Equation


     Before


    Change


     After
modelinginstruction.org

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Before, Change, After (BCA) Tables for Stoichiometry

  • 1.
  • 2. Typical Approach to Stoichiometry  Very algorithmic  grams A --> moles A--> moles B--> grams B  Based on factor-label, unit cancelling, dimensional analysis  Fosters “plug-n-chug” solution  Approach can be used without much conceptual understanding  Disconnected from balanced equation and physical reaction  Relies exclusively on computation ability and favors math-strong students  Especially poor for limiting reactant problems
  • 3. BCA Approach  Stresses mole relationships based on coefficients in balanced chemical equation  Allows students to reason through stoichiometry  Sets up students for equilibrium calculations later (ICE tables)  Clearly shows limiting reactants
  • 4. Emphasis on balanced equation  Step 1- Balance the equation Hydrogen sulfide gas, which smells like rotten eggs, burns in air to produce sulfur dioxide and water.  How many moles of oxygen gas would be needed to completely burn 2.4 moles of hydrogen sulfide? 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before Change After
  • 5. Focus on mole relationships  Step 2: Fill in the “Before” line  Assume more than enough O2 to react 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before 2.4 XS 0 0 Change After
  • 6. Focus on mole relationships  Step 3: Use ratio of coefficients to determine change  This is done using “for every” statement proportional reasoning statements  E.g., according to the reaction, “for every 2mol of hydrogen sulfide, 3mol of oxygen is required”  Reactants are consumed (-) and products are formed (+) 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before 2.4 XS 0 0 Change -2.4 -3.6 +2.4 +2.4 After
  • 7. Emphasize that change and after are not equivalent  Step 4: Complete the table 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before 2.4 XS 0 0 Change -2.4 -3.6 +2.4 +2.4 After 0 XS 2.4 2.4
  • 8. Complete other calculations on the side  In this case, desired answer is in moles  If mass is required, calculate from moles to grams in your usual way 32 .0 γ 3.6 µολεσ 2 × Ο =115 γ 1 µολε
  • 9. Only moles go in the BCA table  The balanced equation deals with: how many, not how much  If given mass of reactants:  Calculate to moles first, then use the table  If asked for mass of products:  Use table first, then calculate to grams
  • 10. Limiting Reactant Problems  BCA approach distinguishes between what you start with and what reacts When 0.50 mol of aluminum reacts with 0.72 mol of iodine to form aluminum iodide, -How many moles of the excess reactant will remain? -How many moles of aluminum iodide will be formed? 2 Al + 3 I2 ----> 2 AlI3 Before 0.5 0.72 0 Change After
  • 11. Limiting Reactant Problems  Assume first reactant is all used, then reason how many moles of the other reactant you would need and compare to what you have:  For every 2 mol of aluminum, you need 3 mol of iodine; for 0.5mol of aluminum, you’d need 0.75 mol of iodine  Since you only have 0.72 mol of iodine, you “don’t have enough” and the iodine will limit the products  Therefore, iodine is the limiting reactant and is totally used up in the reaction 2 Al + 3 I2 ----> 2 AlI3 Before 0.5 0.72 0 Change -0.5 -0.75 After  It is thus clear to students that there’s not enough I2 to react with all the Al
  • 12. Limiting Reactant Problems  Now, proceed with BCA table calculation based on iodine being totally used, and determine the desired quantity(ies):  0.02mol of aluminum are left (could calculate grams if needed)  0.48mol of aluminum iodide are formed (could calculate grams) 2 Al + 3 I2 ----> 2 AlI3 Before 0.5 0.72 0 Change -0.48 -0.72 +0.48 After 0.02 0 0.48
  • 13. BCA Table = Versatile Tool  It doesn’t matter what are the units of the initial given quantities in a stoichiometry problem:  Mass - use molar mass  Gas volume - use molar volume 1 µολε 1.6 γ Ο2 × = 0.050 µολεσ  Solution volume - use molarity 32 .0 γ 1 µολε 7.84 Λ× = 0.350 µολεσ 22 .4 Λ 0.100 µολε 0.0250 Λ× = 0.00250 µολε 1.0 Λ  Calculate to find moles, then use the BCA table  Solve for how many first, and then for how much
  • 14. BCA Table Template Balanced Equation Before Change After