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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 - PowerPoint PPT Presentation
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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
IC
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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C
E
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 12.0 mol 12.0 mol 0
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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C
E
12.0 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C
E
12.0mol O2 x 2 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C
E
12.0 mol O2 x 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C
E
12.0 mol O2 x 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C
E
12.0 mol O2 x 2 mol Ca = 24.0 mol Ca
1 mol O2
Don’t have enough Ca !
12.0 mol Ca x 1 mol O2 = 6.00 mol of O2
2 mol Ca
Yahooooo!
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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C 12.0 mol 6.0 mol
E
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 12.0 mol 12.0 mol 0
C 12.0 mol 6.0 mol
E
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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C 12.0 mol 6.0 mol
E
12.0mol 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C 12.0 mol 6.0 mol
E
12.0mol Ca x 2 mol CaO
2 mol 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C 12.0 mol 6.0 mol
E
12.0mol Ca x 2 mol CaO = 12.0 mol CaO
2 mol 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C 12.0 mol 6.0 mol 12.0 mol
E
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 12.0 mol 12.0 mol 0
C -12.0 mol -6.0 mol +12.0 mol
E
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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C -12.0 mol -6.0 mol +12.0 mol
E 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C -12.0 mol -6.0 mol +12.0 mol
E 0 mol 6.0 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C -12.0 mol -6.0 mol +12.0 mol
E 0 mol 6.0 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.
2 Ca + 1 O2 2 CaO
I 12.0 mol 12.0 mol 0
C -12.0 mol -6.0 mol +12.0 mol
E 0 mol 6.0 mol 12.0 mol
Limiting
Reactant
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 12.0 mol 12.0 mol 0
C -12.0 mol -6.0 mol +12.0 mol
E 0 6.0 mol 12.0 mol
Limiting Excess
Reactant 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.
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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C
E
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 24.0 mol 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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C
E
24.0 Br2 x
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 24.0 mol 24.0 mol 0 mol
C
E
24.0 Br2 x 2 mol Al
3 mol Br2
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 24.0 mol 24.0 mol 0 mol
C
E
24.0 Br2 x 2 mol Al = 16 mol Al
3 mol Br2
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 24.0 mol 24.0 mol 0 mol
C
E
24.0 Br2 x 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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C 16.0 mol 24.0 mol
E
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 24.0 mol 24.0 mol 0 mol
C 16.0 mol 24.0 mol
E
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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C 16.0 mol 24.0 mol
E
24.0 mol Br2
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 24.0 mol 24.0 mol 0 mol
C 16.0 mol 24.0 mol
E
24.0 mol Br2 x 2 mol AlBr3 =
3 mol Br2
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 24.0 mol 24.0 mol 0 mol
C 16.0 mol 24.0 mol
E
24.0 mol Br2 x 2 mol AlBr3 = 16 mol AlBr3
3 mol Br2
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 24.0 mol 24.0 mol 0 mol
C 16.0 mol 24.0 mol 16.0mol
E
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 24.0 mol 24.0 mol 0 mol
C -16.0 mol -24.0 mol +16.0mol
E
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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C -16.0 mol -24.0 mol +16.0mol
E 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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C -16.0 mol -24.0 mol +16.0mol
E 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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C -16.0 mol - 24.0 mol +16.0mol
E 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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C -16.0 mol -24.0 mol +16.0mol
E 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.
2 Al + 3 Br2 2 AlBr3
I 24.0 mol 24.0 mol 0 mol
C -16.0 mol - 24.0 mol +16.0mol
E 8.0 mol 0 mol 16.0mol
Excess Limiting Reactant 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 + 5 O2 2 P2O5
I
C
E
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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
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 14.0mol 4.0mol 0mol
C
E
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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
14.0 mol P4 x
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 14.0mol 4.0mol 0mol
C
E
14.0 mol P4 x 5 mol O2
1 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
14.0 mol P4 x 5 mol O2 = 70 mol O2
1 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0molCE
14.0 mol P4 x 5 mol O2 = 70 mol O2
1 mol P4
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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
4.0 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
4.0 mol O2 x 1 mol P4
5.0 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
4.0 mol O2 x 1 mol P4 = 0.80 mol P4
5.0 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C
E
4.0 mol O2 x 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C 0.80 mol 4.0 mol
E
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 14.0mol 4.0mol 0mol
C 0.80 mol 4.0 mol
E
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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C 0.80 mol 4.0 mol
E
4.0 mol O2 x
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 14.0mol 4.0mol 0mol
C 0.80 mol 4.0 mol
E
4.0 mol O2 x 2 mol P2O5
5.0 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C 0.80 mol 4.0 mol
E
4.0 mol O2 x 2 mol P2O5 = 1.6 mol P2O5
5.0 mol 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C 0.80 mol 4.0 mol 1.6 mol
E
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 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E
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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E 13.2 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E 13.2 mol 0 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E 13.2 mol 0 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E 13.2 mol 0 mol 1.6 mol
Excess 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E 13.2 mol 0 mol 1.6 mol
Excess Limiting Reactant 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.
P4 + 5 O2 2 P2O5
I 14.0mol 4.0mol 0mol
C -0.80 mol -4.0 mol +1.6 mol
E 13.2 mol 0 mol 1.6 mol
Excess Limiting Reactant Reactant
HOMEWORK: WORKSHEET # 6
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) 2Al(OH)3 (s) + 3H2 (g)
Mass 79.12g 185.0g 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) 2Al(OH)3 (s) + 3H2 (g)
Mass 79.12g 185.0g 0g 0gICE
Mass
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) 2Al(OH)3 (s) + 3H2 (g)
I 2.93 mol 10.27 mol 0 mol 0 mol
C
E
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) 2Al(OH)3 (s) + 3H2 (g)
I 2.93 mol 10.27 mol 0 mol 0 mol
C
E
2.93 mol Al x 6 mol H20 = 8.74 mol H2O We have enough!
2 mol Al
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 0 mol
C
E
2.93 mol Al x 6 mol H20 = 8.74 mol
2 mol Al
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 0 mol
C 2.93 mol 8.74 mol
E
2.93 mol Al x 2Al(OH)3 = 2.93 mol Al(OH)3
2 mol Al
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 0 mol
C -2.93 mol -8.74 mol +2.93 mol
E 0 mol 1.53 mol 2.93 mol
Excess
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 0 mol
C -2.93 mol -8.74 mol +2.93 mol
E 0 mol 1.53 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
