Transcript Limiting Reactants and ICE Charts
Limiting Reactants and ICE Charts
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?
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?
I nitial C hange 12 E nd 3 20 8 F + 2 S + 8 8 0 2 M 30 8 22 + 6 E → 1 Cake 48 24 24
The sugar runs out-
limiting ingredient 8 cups
of sugar will make
4 cakes
- so multiple the
recipe by 4
All other ingredients are in excess
0 4 4
Limiting Reactant Problems
14.0 mole Ga and 12.0 mole O 2 react. Find the limiting reactant, the mass of excess reactant and product made.
4 Ga + 3 O 2 → Ga 2 O 3
I
14.0
12.0
0
4
C
16.0
12.0
3 Guess
that one of the reactants runs out- then
check
Cannot consume
16.0 moles
– we only have
14 moles
14.0 mole Ga and 12.0 mole O 2 react. Find the limiting reactant, the mass of excess reactant and product made.
4 Ga + 3 O 2 → Ga 2 O 3
I
14.0
12.0
0
3 2
C
14.0
10.5
7.00
E
0
4
1.5
3
x 32.0 g 1 mole 7.00
x 187.4 g 1 mole
48. g 1.31 x 10 3 g
limiting excess product
The other reactant must
run out-
calculate the
change Subtract
to get what is left- Convert back to
grams
14.0 g of Al reacts with 94.0 g of Br 2 . Find the limiting reactant, the mass of the excess reactant and product.
I C
2
Al + 14.0 g x 1 mole 27.0 g 0.5185 mole 0.5185 mole
3 2 3
Br 2 94.0 g x 1 mole 159.8 g → AlBr 3 0.5882 mole 0.7778 mole 0 mole
I C E 14.0 g of Al reacts with 94.0 g of Br 2 . Find the limiting reactant, the mass of the excess reactant and product.
2
Al + 14.0 g x 1 mole 27.0 g 0.5185 mole
2
0.3921 mole 0.1264 mole
3
x 27.0 g 1 mole
3.41 g 3
Br 2 94.0 g x 1 mole 159.8 g → AlBr 3 0.5882 mole 0.5882 mole 0.0000
limiting 2 3
0 mole 0.3921 mole 0.3921 mole x 266.7 g 1 mole
105 g
25.0 g of H 3 PO 4
reacts with
94.0 g of Ca(NO 3 ) 2
. Find the limiting reactant, the mass of the excess reactant and product.
2
H 3 PO 4 + Ca(NO 3 ) 2 →
1
Ca 3 (PO 4 ) 2 +
25.0 g x 1 mole 98.03 g 94.0 g x 1 mole 164.1 g 6
HNO 3 I 0.2550 mole C 0.2550 mole E 0 mole
3 2
0.5728 mole 0.3825 mole 0.1903 mole
1 3
0 mole 0.1275 mole
6 1
0.1275 mole 0 mole 0.765 mole 0.765 mole
x 164.1 g x 310.3 g x 63.01 g 1 mole 1 mole 1 mole limiting = 31.2 g = 39.6 g = 48.2 g