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Using S ’ mores as a Model

Using S ’ mores as a Model. To Understand Limiting Reactants. What about the elemental state of the ingredients?. We ’ ll assume all Graham crackers come as two, diatomic. Gc 2 Marshmallows are monatomic. M Hershey ’ s are a very large molecule. H 12. Hershey. Hershey. Hershey. Hershey.

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Using S ’ mores as a Model

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  1. 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. Hershey Hershey Hershey Hershey Hershey Hershey How will we define out s’more molecule? • 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:

  4. Hershey Hershey Hershey Hershey Hershey Hershey Hershey Hershey Hershey Hershey Hershey Question #1 Balanced Equation • Gc2 + M + H12 → Gc4MH3 • 8Gc2 + 4 M + H12 → 4Gc4MH3

  5. Working the s’more recipe 8Gc2 + 4 M + H12 → 4Gc4MH3 • With only 3 Hershey’s bars and all the other ingredients that you need, how many s’mores can you make? • How many graham cracker do you need to use those 3 hershey bars? • How many marshmallows would be needed to use all those Hershey bars?

  6. #3 8Gc2 + 4 M + H12 → 4Gc4MH3 • How many s’mores can be made with 5 marshmallows? • How many s’mores can be made with 1 Hershey molecule? • How many s’mores can be made with 6 Graham crackers?

  7. #3 8Gc2 + 4 M + H12 → 4Gc4MH3 • Obviously we max out with 3 s’mores, • thus it’s the graham crackers that limit. • thus 5 M’s started − 3 needed = 2 M left over • thus 1 Hershey started − 0.75 Hershey needed = 0.25 H12 left over (which is only 3 little pieces)

  8. #4 8Gc2 + 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? • Divide each item by its coefficient, and the smallest result will be the limiting reactant. • So Hershey Limits • How many s’mores can be made? • 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.

  9. 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

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