The BCA Method in Stoichiometry
The BCA Method in Stoichiometry
Larry Dukerich
Modeling Instruction Program
Arizona State University
Balanced Chemical Equations
1. Atoms are conserved in chemical reactions: balanced chemical equations.
a. Reactants --> Products b. Ratio of particles: mole ratios through coefficients
• Ex. 2 Mg + 1 O
2
--> 2MgO
• two moles of magnesium burn with one mole of oxygen gas to produce two moles of
MgO
• 2 moles Mg: 1 mole O
2
: 2 moles MgO
Making Predictions
1. Using mole ratios we can identify: a. How much is needed (before) b. How much can be made (after)
2. To make predictions we need also to consider the changes that occur during the reaction.
3. 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 H
2
S + 3 O
2
----> 2 SO
2
+ 2 H
2
O
Before:
Change
After
Focus on mole relationships
• Step 2: fill in the before line
2 H
2
S + 3 O
2
----> 2 SO
2
+ 2 H
2
O
Before:
Change
2.4 xs 0 0
After
• Assume more than enough O
2 to react
Focus on mole relationships
• Step 3: use ratio of coefficients to determine change
2 H
2
S + 3 O
2
----> 2 SO
2
+ 2 H
2
O
Before:
Change
2.4 xs 0 0
–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 H
2
S + 3 O
2
----> 2 SO
2
+ 2 H
2
O
Before:
Change
After
2.4 xs 0 0
–2.4
–3.6
+2.4
+2.4
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
3 .
6 moles O
2
32 .
0 g
1 mole
115 g
Mole Ratios: #1
• Lead will react with hydrochloric acid to produce lead chloride
(PbCl
2
) 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 --> __PbCl
2
+ __H
2
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 --> __PbCl
2
+ __H
2
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 --> __PbCl
2
+ __H
2
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 --> __PbCl
2
+ __H
2
Before: 4 moles excess 0 moles 0 moles
Change: -4 moles -8 moles + 4 mole + 4 mole
____________________________________________________
After: 0 mole excess 4 moles
You Can Convert: moles to grams
4 moles
(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?
CaH
2
+ 2 H
2
O --> Ca(OH)
2
+ 2 H
2
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?
CaH
2
+ 2 H
2
O --> Ca(OH)
2
+ 2 H
2
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?
CaH
2
+ 2 H
2
O --> Ca(OH)
2
+ 2 H
2
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?
CaH
2
+ 2 H
2
O --> Ca(OH)
2
+ 2 H
2
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 C
6
H
6
+ 15 O
2
--> 12 CO
2
+ 6 H
2
O
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 C
6
H
6
+ 15 O
2
--> 12 CO
2
+ 6 H
2
O
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 C
6
H
6
+ 15 O
2
--> 12 CO
2
+ 6 H
2
O
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 C
6
H
6
+ 15 O
2
--> 12 CO
2
+ 6 H
2
O
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 I
2
----> 2 AlI
3
Before: 0.50 0.72 0
Change
After
Limiting reactant problems
• Guess which reactant is used up first, then check
2 Al + 3 I
2
----> 2 AlI
3
Before: 0.50 0.72 0
Change -0.50
-0.75
After
• It’s clear to students that there’s not enough I
2 the Al.
to react with all
Limiting reactant problems
• Now that you have determined the limiting reactant, complete the table, then solve for the desired answer.
2 Al + 3 I
2
----> 2 AlI
3
Before: 0.50 0.72 0
Change -0.48
-0.72
After 0.02
0
+0.48
0.48
BCA a versatile tool
• It doesn’t matter what are the units of the initial quantities
– Mass - use molar mass
1 .
6 g O
2
1 mole
32 .
0 g
0 .
050 moles
– Gas volume - use molar volume
– Solution volume - use molarity
7 .
84 L
0 .
0250 L
1 mole
22 .
4 L
1 .
0 L
0 .
350 moles
0 .
100 mole
• Convert to moles, then use the BCA table
0 .
00250 mole
• Solve for how many, then for how much
