The BCA Method in Stoichiometry Larry Dukerich Modeling Instruction Program Arizona State University

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

BCA Approach Stresses mole relationships based on coefficients in balanced chemical equation Sets up equilibrium calculations later (ICE tables) Clearly shows limiting reactants

Balanced Chemical Equations Atoms are conserved in chemical reactions: balanced chemical equations. Reactants --> Products Ratio of particles: mole ratios through coefficients Ex. 2 Mg + 1 O2 --> 2MgO two moles of magnesium burn with one mole of oxygen gas to produce two moles of MgO 2 moles Mg: 1 mole O2: 2 moles MgO

Making Predictions Using mole ratios we can identify: How much is needed (before) How much can be made (after) To make predictions we need also to consider the changes that occur during the reaction. BCA: Before-Change-After chart to keep track of what happens to particle/mole ratios during a chemical reaction

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

Focus on mole relationships Step 2: fill in the before line 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before: 2.4 xs 0 0 Change After Assume more than enough O2 to react

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 0 Change –2.4 –3.6 +2.4 +2.4 After Reactants are consumed (-), products accumulate (+)

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 0 Change –2.4 –3.6 +2.4 +2.4 After 0 xs 2.4 2.4

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

Mole Ratios: #1 Lead will react with hydrochloric acid to produce lead chloride (PbCl2) and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: Before: Change: ____________________________________________________ After:

Equation: Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + __ HCl --> __PbCl2 + __H2 Before: Change: ____________________________________________________ After:

Balance & Before: Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + _2_ HCl --> __PbCl2 + __H2 Before: 4 moles excess 0 moles 0 moles Change: ____________________________________________________ After:

Change Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + _2_ HCl --> __PbCl2 + __H2 Before: 4 moles excess 0 moles 0 moles Change: -4 moles -8moles + 4 mole + 4 mole ____________________________________________________ After:

After Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + _2_ HCl --> __PbCl2 + __H2 Before: 4 moles excess 0 moles 0 moles Change: -4 moles -8 moles + 4 mole + 4 mole ____________________________________________________ After: 0 mole excess 4 moles 4 moles You Can Convert: moles to grams (multiply moles x molar mass (g/mol))

Mole Rations: #2 How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH2 + 2 H2O --> Ca(OH)2 + 2 H2 Before Change ___________________________________________ After

Before: How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH2 + 2 H2O --> Ca(OH)2 + 2 H2 Before 2.5 moles excess 0 moles 0 moles Change ___________________________________________ After

Change: How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH2 + 2 H2O --> Ca(OH)2 + 2 H2 Before 2.5 moles excess 0 moles 0 moles Change -2.5 moles - 5 moles + 2.5 moles + 5 moles ___________________________________________ After

After How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH2 + 2 H2O --> Ca(OH)2 + 2 H2 Before 2.5 moles excess 0 moles 0 moles Change -2.5 moles - 5 moles + 2.5 moles + 5 moles ___________________________________________ After 0 moles excess 2.5 moles 5 moles

Mole Ratios: #3 How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O Before Change ____________________________________________________ After

Before How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O Before 0.06 mol excess 0 mol 0 mol Change ____________________________________________________ After

Change How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O Before 0.06 mol excess 0 mol 0 mol Change -0.06 mol -0.45 mol + 0.36 mol + 0.18 mol ____________________________________________________ After

After How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O Before 0.06 mol excess 0 mol 0 mol Change -0.06 mol -0.45 mol + 0.36 mol + 0.18 mol ____________________________________________________ After 0 mol excess 0.36 mol 0.18 mol

Only moles go in the BCA table If given mass of reactants for products, convert to moles first, then use the table.

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 The balanced equation deals with how many, not how much. 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 0 Change After

Limiting reactant problems Guess which reactant is used up first, then check 2 Al + 3 I2 ----> 2 AlI3 Before: 0.50 0.72 0 Change -0.50 -0.75 After It’s clear to students that there’s not enough I2 to react with all the Al.

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 0 Change -0.48 -0.72 +0.48 After 0.02 0 0.48

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