Using S’mores as a Model

To Understand Limiting Reactants

2

What about the elemental state of the ingredients?

• We’ll assume all Graham crackers come as two, diatomic.✓ Gc2

• Marshmallows are monatomic.✓ M

• Hershey’s are a very large molecule.✓ H12

3

• Of course a true s’more is really a mixture, but in this model we are calling it a molecule with a fixed ratio of atoms.

• Chemical Formula: Gc4MH3

• Let’s write a balanced equation:

How will we define out s’more molecule?

Hersh

ey

Hersh

ey

Hersh

ey

Hersh

ey

Hersh

ey

Hersh

ey

4

•

• Gc2 + M + H12 → Gc4MH3

• 8 Gc2 + 4 M + H12 → 4 Gc4MH3

Question #1 Balanced Equation

Hersh

ey

Hersh

ey

Hersh

ey

Her

shey

Her

shey

Her

shey

Her

shey

Her

shey

Her

shey

Her

shey

Her

shey

5

Working the s’more recipe 8 Gc2 + 4 M + H12 → 4 Gc4MH3

2.With only 3 Hershey’s bars and all the other ingredients that you need, how many s’mores can you make?

a.

✓ How many graham cracker do you need to use those 3 hershey bars?

b.

✓ How many marshmallows would be needed to use all those Hershey bars?

c.

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#3 8 Gc2 + 4 M + H12 → 4 Gc4MH3

a. How many s’mores can be made with 5 marshmallows?

✓

b. How many s’mores can be made with 1 Hershey molecule?

✓

c. How many s’mores can be made with 6 Graham crackers?

✓

7

#3 8 Gc2 + 4 M + H12 → 4 Gc4MH3

d.Obviously we max out with 3 s’mores, e.thus it’s the graham crackers that limit.

f.

✓thus 5 M’s started − 3 needed = 2 M left over

g.

✓thus 1 Hershey started − 0.75 Hershey needed = 0.25 H12 left over (which is only 3 little pieces)

8

#4 8 Gc2 + 4 M + H12 → 4Gc2MH3• Suppose you were given 50 marshmallows, 5 Hershey bars and 76

graham crackers, Which ingredient limits the amount of s’mores that can be made?

a. Divide each item by its coefficient, and the smallest result will be the limiting reactant.

» » So Hershey Limits

b. How many s’mores can be made?✓

c. How much of each item is left over?✓

✴ 76 Gc2 given − 40 Gc2 needed = 36 Gc2 left over

✓

✴ 50 marshmallows given − 20 needed = 30 marshmallows left over.

OK, So I can understand s’mores, what about with real

chemicals and equations

Limiting Reactant Problems

10

4 Al + 3 O2 2 Al2O3

• If you were given 15 moles of aluminum, and 13 moles of oxygen gas, which substance limits the reaction by running out first?

• 15/4 = 3.75 13/3 = 4.3 therefore aluminum limits

• Which substance is left over and how much?

• 15 moles Al * (3 O2/4Al) = 11.25 moles O2 needed

• 13 - 11.25 = 1.75 moles O2 left over

• How much product can be produced?

• 15 moles Al * (2 Al2O3/4Al) = 7.5 moles of Al2O3 can be

produced

11

5 C + 2 SO2 CS2 + 4 CO• If you had 50.0 g of C and reacted it with 100.0 g of SO2 in the lab,

which reactant limits the reaction?

• 50 g C * (1 mole/12 g) = 4.17 moles C/5 = 0.83

• 100 g SO2 * (1 mole/64 g) = 1.56 moles SO2/2 = 0.78

• Therefore SO2 is the limiting reactant

• Which reactant is left over? What mass?

• 100 g SO2 * (1 mole/64 g) * (5 C/2 SO2) * (12g/1mole) = 46.9 g C needed

• 50 g C given - 46.9 g C needed, therefore 3.1 g C left over.

• What mass of CO can be produced?

• 100 g SO2 * (1 mole/64 g) * (4 CO/2 SO2) * (28g/1mole) = 87.5 g CO can be

produced (theoretically)

• In the Lab you were able to produce 73.5 g of CO, what is the % yield of CO?

• (73.5 g / 87.5 g ) * 100 = 84 % yield CO