The BCA Method in Stoichiometry

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

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 8 moles of hydrogen sulfide?

__ H2S + __ O2 ----> __ SO2 + __ H2O

Before:

Change

After

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 8 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: 8 xs 0 0

Change

After

• Assume more than enough O2 to react

• Products are always ZERO before the reaction starts

Focus on mole relationships

• Step 3: Set up and solve your proportions

2 H2S + 3 O2 ----> 2 SO2 + 2 H2O

Before: 8 xs 0 0

Change X Y Z

After

• Put a variable in for every unknown.

Use a proportion to solve for the unknown value!

Focus on mole relationships• Step 3: Set up and solve your proportions

H2S O2 SO2 H2O

2 3 2 2

8 X Y Z

2 3 2 2 2 2

8 X 8 Y 8 Z

2X = 24 2Y = 16 2Z = 16

X = 12 Y = 8 Z = 8== =

Focus on mole relationships

• Step 3: use ratio of coefficients to determine change

2 H2S + 3 O2 ----> 2 SO2 + 2 H2O

Before: 8 xs 0 0

Change 8 12 8 8

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: 8 xs 0 0

Change 8 12 8 8

After 0 xs 8 8

After can be calculated by subtracting Change values from Before.

One reactant must COMPLETELY be consumed. You must figure out which one that is!

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

gmole

gOmoles 384

1

0.3212 2

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.

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 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

1.6g O2 1mole

32.0g0.050moles

7.84L1mole

22.4L0.350moles

0.0250L0.100mole

1.0L0.00250mole