DO NOW: What mass of aluminum hydroxide is needed to

decompose in order to produce aluminum oxide and 35 L of water at STP?

Word equation:

Formula equation:

After today you will be able to…

• Calculate the limiting reactant using two mass-mass 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 mass-mass calculations will be used.•The L.R. is the one that yields the smaller amount.

Limiting Reactant Example

A real-world example: Making Funfetti Cupcakes!

___mix + ___eggs + ___oil ___pan of cupcakes

Let’s say you have…

1 3 2 1

3 mixes

15 eggs

8 tbsp oil

x

x

x

1 mix1

pan=

=3

eggs

1 pan

2 tbsp oil

1 pan

=

3 pans

5 pans

4 pans

Mixes are the “limiting reactant” because they

are used up first!

Ok… Time to relate this back

to chemistry

…

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 + F2 2KF17.1gK

17.1g K1

x2 mol KF2 mol K

= 25.4g KF 1 mol K39.10g K

x x58.10gKF1 mol KF

14.3gF2 ?gKF1K=39.101F=19.00

58.10g

14.3gF2

12 mol KF1 mol F2

= 43.7g KF 1 mol F2

38.00gF2

x

58.10gKF1 mol KF

x x x

Potassium is the L.R.25.4g of potassium fluoride can be

produced.

K: 17.1gKU: ?gKF

K: 14.3gF2U: ?gKF

After today you will be able to…

• Calculate the limiting reactant using other mole relationships• Calculate how much of the

excess reactant reacts and how much is left over

Other L.R. CalculationsExample: 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?Word equation:Formula Equation:

iron + oxygen iron (III) oxideFe

+ O2 Fe2O33 24

Fe+3 O-2

13.17 mol Fe1

= 6.585 mol Fe2O3

2 mol Fe2O3

4 mol Fe

x

18.19 mol O2

1

= 12.13 mol Fe2O3

2 mol Fe2O3

3 mol O2

x

Can be produced:

K: 13.17 molFeU: ?

molFe2O3

K: 18.19 molO2

U: ? molFe2O3

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?

2Co + 3Cl2 2CoCl33.478x1023atomsCo1

x2 mol CoCl3

2 mol Co=

95.40gCoCl3

1 mol Co

6.02x1023atomsCo

x 165.13gCoCl3

1 mol CoCl3

1Co=58.783Cl=106.35165.13

g

284.7gCoCl3

x

57.92L Cl2

1x2 mol CoCl3

3 mol Cl2

=1 mol Cl2

22.4L Cl2

x 165.13gCoCl3

1 mol CoCl3

x

LR

ERCan be

produced:

K: 3.478x1023 atomsCoU: ? gCoCl3

K: 57.92 LCl2U: ? gCoCl3

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?

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.

19.41L Cl2

3.478x1023atomsCo1

x 3 mol Cl2

2 mol Co= 1 mol Co

6.02x1023atomsCo

x 22.4L Cl2

1 mol CoCl3

x

LR ER

K: 3.478x1023atomsCoU: ? L Cl2

Reacts:

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?

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).

38.51L Cl2

LR ER

57.92L Cl2-19.41L

Cl2Left over:

Complete Worksheet!