Transcript Slide 1

Unit 10:
Stoichiometry
Limiting reactant calculations
Limiting Reactant Calculations
The limiting reactant (L.R.) is the
reactant which runs out first and limits
the amount of product that can be
made.
•Two calculations will be used.
•The L.R. is the one that yields the smaller
amount.
A real-world example: Making
Funfetti Cupcakes!
Ok… Time
Mixes are the “limiting
to
relate
reactant” because they
are used up first!
this back to
chemistry…
1
3
2
1
___mix
+ ___eggs
+ ___oil
 ___pan
of cupcakes
Let’s say you have…
1 pan
= 3 pans
3 mixes x
1 mix
1 pan
15 eggs x
= 5 pans
3 eggs
1 pan
= 4 pans
8 tbsp oil x
2 tbsp oil
Example: If 17.1g of potassium reacts with 14.3g of
fluorine, which reactant is the limiting reactant and what
mass of potassium fluoride can theoretically be produced?
Word equation: potassium + fluorine  potassium fluoride
Formula Equation: 2K
K: 17.1gK
U: ?gKF
17.1gK
+
F2
14.3gF2

17.1g K x 1 mol K x 2 mol KF x 58.10gKF
=
39.10g K
1
2 mol K 1 mol KF
K: 14.3gF2
U: ?gKF
14.3gF2 x 1 mol F2 x 2 mol KF x 58.10gKF =
38.00gF2 1 mol F2 1 mol KF
1
2KF
?gKF
1K=39.10
1F=19.00
58.10g
25.4g KF
43.7g KF
Potassium is the L.R.
25.4g of potassium fluoride can be produced.
Unit 10:
Stoichiometry
Excess reactant calculations
Other L.R. Calculations
Example: How many moles of iron (III) oxide can be
produced from the reaction of 13.17 moles of iron with
18.19 moles of oxygen?
Fe+3
O-2
Word equation:
iron + oxygen  iron (III) oxide
Formula Equation: 4 Fe + 3 O2  2 Fe2O3
K: 13.17 molFe
U: ? molFe2O3
Can be produced:
13.17 mol Fe x 2 mol Fe2O3
= 6.585 mol Fe2O3
4 mol Fe
1
K: 18.19 molO
U: ? molFe2O3
2
18.19 mol O2 x 2 mol Fe2O3 =
12.13 mol Fe2O3
3
mol
O
1
2
Example: Limiting Reactant AND Excess
Reactant Calculations
a) What mass of CoCl3 is formed from the reaction of
3.478x1023 atoms Co with 57.92L of Cl2 gas at STP?
b) How much of the excess reactant reacts and how much
is left over?
1Co=58.78
K: 3.478x1023 atomsCo
U: ? gCoCl3
LR
2Co + 3Cl2  2CoCl3
3Cl=106.35
165.13g
1 mol Co x 2 mol CoCl3 x 165.13gCoCl3
3.478x1023atomsCo x
=
1 mol CoCl3
6.02x1023atomsCo 2 mol Co
1
K: 57.92 LCl2
U: ? gCoCl3
ER
Can be produced:
57.92L Cl2 x 1 mol Cl2 x 2 mol CoCl3 x 165.13gCoCl3
=
22.4L Cl2
3 mol Cl2
1
1 mol CoCl3
95.40gCoCl3
284.7gCoCl3
Example: Limiting Reactant AND Excess
Reactant Calculations
a) What mass of CoCl3 is formed from the reaction of
3.478x1023 atoms Co with 57.92L of Cl2 gas at STP?
b) How much of the excess reactant reacts and how much
is left over?
LR
ER
2Co + 3Cl2  2CoCl3
To find out how much excess reactant reacts, do a third calculation using the
limiting reactant as the known, and the excess reactant as the unknown.
K: 3.478x1023atomsCo
U: ? L Cl2
3.478x1023atomsCo x
1 mol Co x 3 mol Cl2 x 22.4L Cl2
=
1 mol Cl2
1
6.02x1023atomsCo 2 mol Co
Reacts:
19.41L Cl2
Example: Limiting Reactant AND Excess
Reactant Calculations
a) What mass of CoCl3 is formed from the reaction of
3.478x1023 atoms Co with 57.92L of Cl2 gas at STP?
b) How much of the excess reactant reacts and how much
is left over?
LR
ER
2Co + 3Cl2  2CoCl3
To find out how much is left over (unreacted), subtract the amount of the
excess reactant that reacted from the original amount of excess reactant
(from the problem).
57.92L Cl2
-19.41L Cl2
Left over:
38.51L Cl2