Limiting Reactants and
ICE Charts
“Chemistry” Salami Sandwiches
Question:
If you are given one dozen loaves of bread, a gallon
of mustard and three pieces of salami, how many
salami sandwiches can you make?
Answer: Give reasons
Chemistry Cake
Chemistry Cake
You have 20 cups of flour, 8 cups of sugar, 30
litres of milk and 48 eggs in your kitchen.
The recipe for chemistry cake is:
Chemistry Cake
You have 20 cups of flour, 8 cups of sugar, 30
litres of milk and 48 eggs in your kitchen.
The recipe for chemistry cake is:
3 cups of flour
2 cups of sugar
2 litres of milk
+
6 eggs
= 1 chemistry cake
1. How many cakes can you make?
1. How many cakes can you make?
2. Which ingredient ran out first and limited
the number of cakes you could make?
1. How many cakes can you make?
2. Which ingredient ran out first and limited
the number of cakes you could make?
3. What and how much of each ingredient is
left over?
1. How many cakes can you make?
2. Which ingredient ran out first and limited
the number of cakes you could make?
3. What and how much of each ingredient is
left over?
4. What does this assignment have to do with
chemistry?
Limiting Reactant Problems
In the chemical reaction:
2 Al
+
3 Br2
2 AlBr3
In the chemical reaction:
2 Al
+
3 Br2
2 AlBr3
In the chemical reaction:
2 Al + 3 Br2
2 AlBr3
2 moles of Al require 3 moles of Br2 to
produce 2 moles of AlBr3
In the chemical reaction
2 Al
+
3 Br2
2 AlBr3
2 moles of Al require 3 moles of Br2 to
produce 2 moles of AlBr3
What if Al and Br2 are not present in a perfect
2 : 3 ratio?
Answer:
Answer:
One reactant runs out:
Answer:
One reactant runs out:
There is a reactant left over:
Answer:
One reactant runs out:
Limiting Reactant
There is a reactant left over:
Answer:
One reactant runs out:
Limiting Reactant
There is a reactant left over: Excess Reactant
Answer:
One reactant runs out:
Limiting Reactant
There is a reactant left over: Excess Reactant
We use an ICE chart to find out how much
reactant in excess is left over and how much
product is produced.
Answer:
One reactant runs out:
Limiting Reactant
There is a reactant left over:
Excess Reactant
We use an ICE chart to find out how much reactant in
excess is left over and how much product is produced.
I = Initial moles (unreacted reactants)
Answer:
One reactant runs out:
Limiting Reactant
There is a reactant left over:
Excess Reactant
We use an ICE chart to find out how much reactant in
excess is left over and how much product is produced.
I = Initial moles (unreacted reactants)
C = Change in moles (reactants consumed & products produced)
Answer:
One reactant runs out:
Limiting Reactant
There is a reactant left over:
Excess Reactant
We use an ICE chart to find out how much reactant in excess
is left over and how much product is produced.
I = Initial moles (unreacted reactants)
C = Change in moles (reactants consumed & products produced)
E = End moles
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
2 Ca
+
1 O2
2 CaO
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
2 Ca +
1 O2
2 CaO
I
C
E
Write down the INITIAL moles given in the question. If
grams given we need to convert to moles.
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
1 O2
12.0 mol
2 CaO
0
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
2 Ca
+
12.0 mol
1 O2
12.0 mol
2 CaO
0
I
C
E
Now CALCULATE the moles of each reactant needed for
the reaction. Start with whichever element you want.
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol O2
1 O2
12.0 mol
2 CaO
0
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0mol O2
x
1 O2
12.0 mol
2 mol Ca
1 mol O2
2 CaO
0
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
2 Ca
+
12.0 mol
I
C
E
12.0 mol O2
x
1 O2
12.0 mol
2 CaO
0
2 mol Ca = 24.0 mol Ca
1 mol O2
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol O2
x
1 O2
12.0 mol
2 CaO
0
2 mol Ca = 24.0 mol Ca
1 mol O2
Don’t have
enough Ca! 
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol O2
x
12.0 mol Ca
x
1 O2
12.0 mol
2 CaO
0
2 mol Ca = 24.0 mol Ca
1 mol O2
Don’t have
enough Ca ! 
1 mol O2 = 6.00 mol of O2Yahooooo! 
2 mol Ca
Enough O2 !!!
This means for all 12 moles of Ca to be used up we need 6 moles of O2.
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol
1 O2
12.0 mol
6.0 mol
2 CaO
0
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol
1 O2
12.0 mol
6.0 mol
2 CaO
0
Since the limiting reagent determines how much product is
formed we use the moles of limiting reagent to determine
the moles and mass of product
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol
12.0mol Ca
1 O2
12.0 mol
6.0 mol
2 CaO
0
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol
12.0mol Ca
x
1 O2
12.0 mol
6.0 mol
2 mol CaO
2 mol Ca
2 CaO
0
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol
12.0mol Ca
x
1 O2
12.0 mol
6.0 mol
2 mol CaO
2 mol Ca
2 CaO
0
= 12.0 mol CaO
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
12.0 mol
1 O2
12.0 mol
6.0 mol
2 CaO
0
12.0 mol
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
-12.0 mol
1 O2
12.0 mol
-6.0 mol
2 CaO
0
+12.0 mol
Now complete the ICE chart by filling in the END amounts of
each species.
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
-12.0 mol
0 mol
1 O2
12.0 mol
-6.0 mol
2 CaO
0
+12.0 mol
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
-12.0 mol
0 mol
1 O2
12.0 mol
-6.0 mol
6.0 mol
2 CaO
0
+12.0 mol
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
-12.0 mol
0 mol
1 O2
12.0 mol
-6.0 mol
6.0 mol
2 CaO
0
+12.0 mol
12.0 mol
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
-12.0 mol
0 mol
Limiting
Reactant
1 O2
12.0 mol
-6.0 mol
6.0 mol
2 CaO
0
+12.0 mol
12.0 mol
12.0 mole Ca and 12.0 mole O2 react. Find
the limiting reactant, the amount of excess
reactant and the amount of product made.
I
C
E
2 Ca
+
12.0 mol
-12.0 mol
0
Limiting
Reactant
1 O2
12.0 mol
-6.0 mol
6.0 mol
Excess
Reactant
2 CaO
0
+12.0 mol
12.0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
2 Al
+
3 Br2
2 AlBr3
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
2 Al
+
3 Br2
2 AlBr3
I
C
E
Write down the INITIAL amounts given in the question
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
2 Al +
24.0 mol
3 Br2
2 AlBr3
24.0 mol
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
2 Al +
24.0 mol
3 Br2
2 AlBr3
I
24.0 mol
0 mol
C
E
Now CALCULATE the moles of each reactant needed for the
reaction. Start with whichever element you want.
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
24.0 Br2 x
2 Al +
24.0 mol
3 Br2
2 AlBr3
24.0 mol
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
24.0 Br2 x
2 Al +
24.0 mol
2 mol Al
3 mol Br2
3 Br2
2 AlBr3
24.0 mol
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
24.0 Br2 x
2 Al +
24.0 mol
2 mol Al
3 mol Br2
3 Br2
2 AlBr3
24.0 mol
= 16 mol Al
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
24.0 Br2 x
2 Al +
24.0 mol
3 Br2
2 AlBr3
24.0 mol
0 mol
2 mol Al
= 16 mol Al
3 mol Br2
This means for all 24.0 moles of Br2 to be used up we need 16
moles of Al.
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
2 Al +
24.0 mol
16.0 mol
3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
2 Al +
24.0 mol
16.0 mol
3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
Since the limiting reagent determines how much product is
formed we use the moles of limiting reagent to determine
the moles and mass of product
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
24.0 mol Br2
2 Al +
24.0 mol
16.0 mol
3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
2 Al +
24.0 mol
16.0 mol
3 Br2
24.0 mol Br2 x 2 mol AlBr3
3 mol Br2
2 AlBr3
24.0 mol
24.0 mol
=
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
I
C
E
2 Al +
24.0 mol
16.0 mol
3 Br2
24.0 mol Br2 x 2 mol AlBr3
3 mol Br2
2 AlBr3
24.0 mol
24.0 mol
=
16 mol AlBr3
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al +
24.0 mol
16.0 mol
3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
16.0mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al + 3 Br2
24.0 mol
24.0 mol
-16.0 mol
-24.0 mol
2 AlBr3
0 mol
+16.0mol
Now complete the ICE chart by filling in the END amounts of
each species
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al + 3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
-16.0 mol
-24.0 mol
+16.0mol
8.0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al + 3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
-16.0 mol
-24.0 mol
+16.0mol
8.0 mol
0 mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al + 3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
-16.0 mol
- 24.0 mol
+16.0mol
8.0 mol
0 mol
16.0mol
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al + 3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
-16.0 mol
-24.0 mol
+16.0mol
8.0 mol
0 mol
16.0mol
Excess
Reactant
24.0 mole of Al reacts with 24.0 mol of Br2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
2 Al + 3 Br2
2 AlBr3
24.0 mol
24.0 mol
0 mol
-16.0 mol
- 24.0 mol
+16.0mol
8.0 mol
0 mol
16.0mol
Excess
Reactant
Limiting
Reactant
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
P4
+
5 O2
2 P2O5
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
P4
I
C
E
+
5 O2
2 P2O5
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount
of product made.
P4
+
5 O2
2 P2O5
I
C
E
Write down the INITIAL amounts given in the question
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
5 O2
4.0mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
5 O2
4.0mol
2 P2O5
0mol
Now CALCULATE the moles of each reactant needed for the
reaction. Start with whichever element you want
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
14.0 mol P4 x
P4
+
14.0mol
5 O2
4.0mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
14.0 mol P4 x
P4
+
14.0mol
5 mol O2
1 mol P4
5 O2
4.0mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
14.0 mol P4 x
P4
+
14.0mol
5 mol O2
1 mol P4
5 O2
4.0mol
= 70 mol O2
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
14.0 mol P4 x
P4
+
14.0mol
5 mol O2
1 mol P4
5 O2
4.0mol
2 P2O5
0mol
= 70 mol O2
That’s more O2 than we have so redo the calculation starting
with O2
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
4.0 mol O2
P4
+
14.0mol
5 O2
4.0mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
4.0 mol O2 x
P4
+
14.0mol
1 mol P4
5.0 mol O2
5 O2
4.0mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
4.0 mol O2 x
P4
+
14.0mol
1 mol P4
5.0 mol O2
5 O2
4.0mol
= 0.80 mol P4
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
4.0 mol O2 x
P4
+
14.0mol
5 O2
4.0mol
2 P2O5
0mol
1 mol P4
= 0.80 mol P4
5.0 mol O2
This means for all 4.0 moles of O2 to be used up we need 0.80
moles of P4.
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
0.80 mol
5 O2
4.0mol
4.0 mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
0.80 mol
5 O2
4.0mol
4.0 mol
2 P2O5
0mol
Since the limiting reagent determines how much product is
formed we use the moles of limiting reagent to determine the
moles and mass of product
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
4.0 mol O2 x
P4
+
14.0mol
0.80 mol
5 O2
4.0mol
4.0 mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
0.80 mol
4.0 mol O2 x 2 mol P2O5
5.0 mol O2
5 O2
4.0mol
4.0 mol
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
4.0 mol O2 x
P4
+
14.0mol
0.80 mol
5 O2
4.0mol
4.0 mol
2 mol P2O5 = 1.6 mol P2O5
5.0 mol O2
2 P2O5
0mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
0.80 mol
5 O2
4.0mol
4.0 mol
2 P2O5
0mol
1.6 mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
-0.80 mol
5 O2
4.0mol
-4.0 mol
2 P2O5
0mol
+1.6 mol
Now complete the ICE chart by filling in the END amounts of
each species
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
-0.80 mol
13.2 mol
5 O2
4.0mol
-4.0 mol
2 P2O5
0mol
+1.6 mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
-0.80 mol
13.2 mol
5 O2
4.0mol
-4.0 mol
0 mol
2 P2O5
0mol
+1.6 mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
-0.80 mol
13.2 mol
5 O2
4.0mol
-4.0 mol
0 mol
2 P2O5
0mol
+1.6 mol
1.6 mol
14.0 moles P4 react with 4.0 moles O2. Find the
limiting reactant, the amount of the excess reactant
and the amount of product made.
I
C
E
P4
+
14.0mol
-0.80 mol
13.2 mol
5 O2
4.0mol
-4.0 mol
0 mol
Excess Reactant
2 P2O5
0mol
+1.6 mol
1.6 mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
-0.80 mol
13.2 mol
5 O2
4.0mol
-4.0 mol
0 mol
Excess
Reactant
Limiting
Reactant
2 P2O5
0mol
+1.6 mol
1.6 mol
14.0 moles P4 react with 4.0 moles O2. Find the limiting
reactant, the amount of the excess reactant and the amount of
product made.
I
C
E
P4
+
14.0mol
-0.80 mol
13.2 mol
5 O2
4.0mol
-4.0 mol
0 mol
Excess
Reactant
Limiting
Reactant
HOMEWORK: WORKSHEET # 6
2 P2O5
0mol
+1.6 mol
1.6 mol
When 79.12g of aluminum is reacted with 185.0g of
hot water (a SR reaction), the products are aluminum
hydroxide and hydrogen gas. Which of the reactants is
in excess? How many grams of aluminum hydroxide
are produced?
__ Al (s) + __ H2O (l)
__Al(OH)3 (s) + __H2 (g)
When 79.12g of aluminum is reacted with 185.0g of
hot water (a SR reaction), the products are aluminum
hydroxide and hydrogen gas. Which of the reactants is
in excess? How many grams of aluminum hydroxide
are produced?
Write and Balance equation!
2 Al (s) + 6H2O (l)
Mass 79.12g 185.0g
2Al(OH)3 (s) + 3H2 (g)
0g
0g
When 79.12g of aluminum is reacted with 185.0g of hot water (a SR
reaction), the products are aluminum hydroxide and hydrogen gas. Which of
the reactants is in excess? How many grams of aluminum hydroxide are
produced?
2 Al (s) + 6H2O (l)
Mass 79.12g
I
C
E
Mass
185.0g
2Al(OH)3 (s) + 3H2 (g)
0g
0g
79.12g Al x 1 mol Al = 2.930 mol Al
27.0 g Al
185.0g H2O x 1 mol H2O = 10.27 mol H2O
18.0 g H2O
When 79.12g of aluminum is reacted with 185.0g of hot water (a
SR reaction), the products are aluminum hydroxide and
hydrogen gas. Which of the reactants is in excess? How many
grams of aluminum hydroxide are produced?
2 Al (s) + 6H2O (l)
I
C
E
2.93 mol 10.27 mol
2Al(OH)3 (s) + 3H2 (g)
0 mol
0 mol
Now CALCULATE the moles of each reactant needed for the
reaction. Start with whichever element you want
When 79.12g of aluminum is reacted with 185.0g of hot water (a SR
reaction), the products are aluminum hydroxide and hydrogen gas.
Which of the reactants is in excess? How many grams of aluminum
hydroxide are produced?
2 Al (s) + 6H2O (l)
I
2Al(OH)3 (s) + 3H2 (g)
2.93 mol 10.27 mol
0 mol
0 mol
C
E
2.93 mol Al x 6 mol H20 = 8.74 mol H2O
2 mol Al
We have enough!
When 79.12g of aluminum is reacted with 185.0g of hot water (a
SR reaction), the products are aluminum hydroxide and hydrogen
gas. Which of the reactants is in excess? How many grams of
aluminum hydroxide are produced?
2 Al (s) + 6H2O (l)
I
2.93 mol 10.27 mol
C
E
2.93 mol Al x 6 mol H20 = 8.74 mol
2 mol Al
2Al(OH)3 (s) + 3H2 (g)
0 mol
0 mol
When 79.12g of aluminum is reacted with 185.0g of hot water (a
SR reaction), the products are aluminum hydroxide and
hydrogen gas. Which of the reactants is in excess? How many
grams of aluminum hydroxide are produced?
2 Al (s) + 6H2O (l)
I
2.93 mol 10.27 mol
C
2.93 mol
2Al(OH)3 (s) + 3H2 (g)
8.74 mol
E
2.93 mol Al x 2Al(OH)3 = 2.93 mol Al(OH)3
2 mol Al
0 mol
0 mol
When 79.12g of aluminum is reacted with 185.0g of hot water (a
SR reaction), the products are aluminum hydroxide and hydrogen
gas. Which of the reactants is in excess? How many grams of
aluminum hydroxide are produced?
2 Al (s) + 6H2O (l)
2Al(OH)3 (s) + 3H2 (g)
I
2.93 mol 10.27 mol
0 mol
C
-2.93 mol -8.74 mol
+2.93 mol
E
0 mol
1.53 mol
Excess
2.93 mol
0 mol
When 79.12g of aluminum is reacted with 185.0g of hot water (a
SR reaction), the products are aluminum hydroxide and hydrogen
gas. Which of the reactants is in excess? How many grams of
aluminum hydroxide are produced?
2 Al (s) + 6H2O (l)
2Al(OH)3 (s) + 3H2 (g)
I
2.93 mol 10.27 mol
0 mol
C
-2.93 mol -8.74 mol
+2.93 mol
E
0 mol
1.53 mol
0 mol
2.93 mol
2.93 mol Al(OH)3 x 78.0 g Al(OH)3 = 228.5 g Al(OH)3
1 mol Al(OH)3
Now do Worksheet # 7