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- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. BCA Table TemplateBalanced Equation Before Change After
- 14. modelinginstruction.org

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