The document discusses two approaches to stoichiometry: a traditional algorithmic method and a more conceptual bca approach that emphasizes understanding mole relationships based on balanced chemical equations. The bca method helps students reason through stoichiometry, particularly in limiting reactant problems, by using a systematic approach to set up calculations and highlighting the importance of coefficients in reactions. Overall, it presents the bca table as a versatile tool for dealing with various initial quantities in stoichiometry calculations.

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

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