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Chemistry 12.1 Stoichiometry

Chemistry 12.1 Stoichiometry. “The arithmetic of equations”. Chapter 12 Objective. This unit will allow you to calculate the amounts of chemical substances involved in a chemical reaction.  You will use balanced chemical equations to do this…. Tiny Tricycles – 12.1 wks… Cookie Example.

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Chemistry 12.1 Stoichiometry

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  1. Chemistry 12.1Stoichiometry “The arithmetic of equations”

  2. Chapter 12 Objective This unit will allow you to calculate the amounts of chemical substances involved in a chemical reaction.  You will use balanced chemical equations to do this…

  3. Tiny Tricycles – 12.1 wks… Cookie Example

  4. B. Cookie Example 1. Recipe… 2. You have… 2 large eggs + 1/2 cup sugar + 1 teaspoon vanilla + 3 tbsp cocoa powder + 1/2 cup chocolate chips  8 chocolate chip cookies 16 large eggs 12 cups sugar 6 teaspoons vanilla 33 tbsp cocoa powder 18 cups chocolate chips  ? chocolate chip cookies 48 chocolate chip cookies

  5. 3. What do you have left? 4. What do you need for 4 dozen cookies? 4 eggs 9 cups of sugar 15 tbsp of cocoa powder 15 cups of chocolate chips 12 eggs 3 cups of sugar 6 teaspoons of vanilla 18 tbsp of cocoa powder 3 cups of chocolate chips

  6. Write you own example equation about and item that has multiple parts/pieces…

  7. I. Stoichiometry A. Def - a calculation of chemical quantities in a chemical reaction. 1. Our formula 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 2. What can we determine with a balanced chemical equation??? a. 1 : 3 : 2 ratio -this applies to atoms, molecules, or moles; mass; and volume.

  8. 1 N2 (g) + 3 H2 (g) 2 NH3 (g) Particles… Moles… Grams… • 2 atoms of N 6 atoms of H 2 atoms N + 6 atoms H Particles are conserved… 1 mol N2 + 3 mol H2  2 mol NH3 Moles don’t have to be conserved… 28 g of N2 + 6 g of H2  34 g of NH3 Mass is conserved…

  9. Gas/Volume (at STP)… *Mass and atoms are always conserved… *Molecules, formula units, moles, and volumes are not necessarily conserved 22.4 L of N2 + 67.2 L of H2  44.8 L of NH3 Volume doesn’t have to be conserved…

  10. 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O(g) Hydrogen sulfide, a foul smelling gas, is found in nature in volcanic areas. The balanced chemical equation for the burning of hydrogen sulfide is given above. Interpret this equation in terms of the interaction of the three relative quantities. Do to this equation as what was on the last screen. a. Number of particles b. Number of moles c. Masses of reactants and products

  11. a. Particles b. Moles c. Grams 2 molecules H2S + 3 molecules O2  2 molecules SO2 + 2 molecules H2O 2 mol H2S + 3 mol O2  2 mol SO2 + 2 mol H2O 68.2 g H2S + 96.0 g O2  128.2 g SO2 + 36.0 g H2O 164.2 g reactants = 164.2 g products

  12. In Class Assignment • C2H2(g) + O2(g) CO2(g) + H2O (g) a. Balance b. Number of particles c. Number of moles d. Masses of reactants and products

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