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Before, Change, After (BCA) Tables for Stoichiometry

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The BCA Table method of performing stoichiometry calculations that is a cognitive approach that does not rely on algorithms, but rather it engages proportional reasoning skills.

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Before, Change, After (BCA) Tables for Stoichiometry

  1. 1. 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
  2. 2. 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
  3. 3. 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
  4. 4. 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
  5. 5. 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
  6. 6. 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
  7. 7. 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
  8. 8. 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
  9. 9. 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
  10. 10. 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
  11. 11. 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
  12. 12. 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
  13. 13. BCA Table TemplateBalanced Equation Before Change After
  14. 14. modelinginstruction.org

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