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Typical Approach to Stoichiometry . Very algorithmicgrams A -->moles A-->moles B-->grams BFosters plug-n-chug solutionDisconnected from balanced equationEspecially poor for limiting reactant problems. BCA Approach. Stresses mole relationships based on coefficients in balanced chemical equationS
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1. The BCA Method in Stoichiometry Larry Dukerich
Modeling Instruction Program
Arizona State University
2. Typical Approach to Stoichiometry Very algorithmic
grams A -->moles A-->moles B-->grams B
Fosters plug-n-chug solution
Disconnected from balanced equation
Especially poor for limiting reactant problems
3. BCA Approach Stresses mole relationships based on coefficients in balanced chemical equation
Sets up 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
2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
Before: 2.4 xs 0 0Change After
Assume more than enough O2 to react
6. Focus on mole relationships Step 3: use ratio of coefficients to determine change
2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
Before: 2.4 xs 0 0Change –2.4 –3.6 +2.4 +2.4 After
Reactants are consumed (-), products accumulate (+)
7. Emphasize that change and final state are not equivalent Step 4: Complete the table
2 H2S + 3 O2 ----> 2 SO2 + 2 H2O
Before: 2.4 xs 0 0Change –2.4 –3.6 +2.4 +2.4 After 0 xs 2.4 2.4
8. Complete calculations on the side In this case, desired answer is in moles
If mass is required, convert moles to grams in the usual way
9. Only moles go in the BCA table The balanced equation deals with how many, not how much.
If given mass of reactants for products, convert to moles first, then use the table.
10. Limiting reactant problems BCA approach distinguishes between what you start with and what reacts.
When 0.50 mole of aluminum reacts with 0.72 mole 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: 0.50 0.72 0Change After
11. Limiting reactant problems Guess which reactant is used up first, then check
2 Al + 3 I2 ----> 2 AlI3
Before: 0.50 0.72 0Change -0.50 -0.75 After
It’s clear to students that there’s not enough I2 to react with all the Al.
12. Limiting reactant problems Now that you have determined the limiting reactant, complete the table, then solve for the desired answer.
2 Al + 3 I2 ----> 2 AlI3
Before: 0.50 0.72 0Change -0.48 -0.72 +0.48 After 0.02 0 0.48
13. BCA a versatile tool It doesn’t matter what are the units of the initial quantities
Mass - use molar mass
Gas volume - use molar volume
Solution volume - use molarity
Convert to moles, then use the BCA table
Solve for how many, then for how much