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. ✓ Gc 2 Marshmallows are monatomic. ✓M✓M Hershey’s are a very large molecule. ✓ H 12

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: Gc 4 MH 3 Let’s write a balanced equation: How will we define out s’more molecule? Hershey

4 Gc 2 + M + H 12 → Gc 4 MH 3 8 Gc M + H 12 → 4 Gc 4 MH 3 Question #1 Balanced Equation Hershey

5 Working the s’more recipe 8 Gc M + H 12 → 4 Gc 4 MH 3 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.

6 #38 Gc M + H 12 → 4 Gc 4 MH 3 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 #38 Gc M + H 12 → 4 Gc 4 MH 3 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 H 12 left over (which is only 3 little pieces)

8 #48 Gc M + H 12 → 4Gc 2 MH 3 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 Gc 2 given − 40 Gc 2 needed = 36 Gc 2 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 O 2  2 Al 2 O 3 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 = 4.3 therefore aluminum limits Which substance is left over and how much? 15 moles Al * (3 O 2 /4Al) = moles O 2 needed = 1.75 moles O 2 left over How much product can be produced? 15 moles Al * (2 Al 2 O 3 /4Al) = 7.5 moles of Al 2 O 3 can be produced

11 5 C + 2 SO 2  CS CO If you had 50.0 g of C and reacted it with g of SO 2 in the lab, which reactant limits the reaction? 50 g C * (1 mole/12 g) = 4.17 moles C/5 = g SO 2 * (1 mole/64 g) = 1.56 moles SO 2 /2 = 0.78 Therefore SO 2 is the limiting reactant Which reactant is left over? What mass? 100 g SO 2 * (1 mole/64 g) * (5 C/2 SO 2 ) * (12g/1mole) = 46.9 g C needed 50 g C given g C needed, therefore 3.1 g C left over. What mass of CO can be produced? 100 g SO 2 * (1 mole/64 g) * (4 CO/2 SO 2 ) * (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