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.
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