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Stoichiometry

Stoichiometry. Simply, the math behind chemistry. What is a chemical equation?. Can you define these?. Reactants Products Balance Law of Conservation of Mass The Mole Mass to Mole conversion Molar Mass. You Tube Crash Course. http://www.youtube.com/watch?v=UL1jmJaUkaQ.

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Stoichiometry

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  1. Stoichiometry Simply, the math behind chemistry.

  2. What is a chemical equation?

  3. Can you define these? • Reactants • Products • Balance • Law of Conservation of Mass • The Mole • Mass to Mole conversion • Molar Mass

  4. You Tube Crash Course • http://www.youtube.com/watch?v=UL1jmJaUkaQ

  5. When you use stoichiometry, you can determine amounts of substances needed to fulfill the requirements of the reaction.

  6. Let's start with something simple like sodium chloride (NaCl). • You start with two ions and wind up with an ionic/electrovalent compound. What type of reaction is this? • When you look at the equation, you see that it takes one sodium ion (Na+) to combine with one chlorine ion (Cl-) to make the salt.

  7. Nothing you can do will change that. 10,000,000 Na + 1 Cl --> 1 NaCl + 9,999,999 Na

  8. Let's bump it up a level. • When you mix hydrogen gas (H2) and oxygen gas (O2), nothing much happens. Why?

  9. Add a spark to the mixture ! • the molecules combine and eventually form water (H2O). You would write it like this: 2H2 + O2 --> 2H2O • What does stoichiometry look at here?

  10. If you make this an extreme example • Fill a sealed container with one million hydrogen molecules • And only one oxygen molecule • Add a spark • What happens?

  11. You won't make an explosion. • There is no monster reaction to be created • There is only one oxygen molecule • You will make two water molecules • And be done.

  12. Limiting Reagents • Sometimes when reactions occur between two or more substances, one reactant runs out before the other. • That is called the "limiting reagent". • Often, it is necessary to identify the limiting reagent in a problem.

  13. Example: A chemist only has 6.0 grams of C2H2 and an unlimited supply of oxygen and he desires to produce as much CO2 as possible. If she uses the equation below, how much oxygen should she add to the reaction?

  14. 2C2H2 (g) + 5O2 (g) ---> 4CO2 (g) + 2 H2O(l) • To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up. • First, we calculate the number of moles in the given reactant. • Second, use mole ratio. • Third convert the other reactant to grams.

  15. First, we calculate the number of moles of C2H2 in 6.0 g of C2H2. • 1 mole of C2H2 weighs 26 g 2 C = 2 × 12 2 H = 2 × 1 • 6.0 g C2H2 x 1 mol C2H2 26g C2H2 = 0.25 mol C2H2

  16. 2C2H2 (g) + 5O2 (g) ---> 4CO2 (g) + 2 H2O(l) • Then, because there are five (5) molecules of oxygen to every two (2) molecules of C2H2, we need to multiply the result by 5/2 to get the total molecules of oxygen. 0.25 mol C2H2 x 5 mol O2 = 0.625 mol O2 2 mol C2H2

  17. Then we convert to grams to find the amount of oxygen that needs to be added: 0.625 mol O2x32.0 g O2 1 mol O2 = 20 g O2

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