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Stoichiometry 1

Stoichiometry 1. Mole-Mole and Mass-Mass Conversions. Review. In chemical reactions, mass is neither created nor destroyed. Law of Conservation of Mass A balanced chemical equation obeys the Law of Conservation of Mass. WRONG: N 2 + H 2  NH 3 RIGHT: N 2 + 3H 2  2NH 3.

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Stoichiometry 1

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  1. Stoichiometry 1 Mole-Mole and Mass-Mass Conversions

  2. Review • In chemical reactions, mass is neither created nor destroyed. • Law of Conservation of Mass • A balanced chemical equation obeys the Law of Conservation of Mass. • WRONG: N2 + H2 NH3 • RIGHT: N2 + 3H2  2NH3

  3. Using Chemical Equations • Chemical equations are like recipes: • They tell chemists what amounts of reactants to mix and what amount of products to expect. • If you know the quantity of one substance in a rxn, you can calculate the quanity of any other substance in the same rxn (reactant or product). • Quantity = amount of a substance in grams, liters, molecules, or moles. • The calculation of quantities in chemical rxns is called stoichiometry.

  4. Stoichiometry • A bicycle has 5 major components: • 1 frame (F) • 1 seat (S) • 2 wheels (2W) • 1 handlebar (H) • 2 pedals (2P) • A complete bicycle has a “formula” of FSW2HP2. • F + S + 2W + H + 2P  FSW2HP2.

  5. + + + + H P F S 2W + + + + FSW HP 2 2 Stoichiometry

  6. Stoichiometry • Suppose you run a bicycle manufacturing company. If you have 500 wheels on stock, how many bicycles can you make? (Assume you have more than enough of the other parts). • 500 wheels x • If you need to make 640 bicycles by the end of the week, how many bicycle seats will you need? • 640 bicycles = 250 bicycles 1 bicycle 1 seat 2 wheels 1 bicycle = 640 seats

  7. Stoichiometry • The coefficients in a balanced chemical equation (hereafter, BCE) are the numbers of moles of reactants and products in a chemical reaction. • N2 + 3H2 2NH3 • 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. • N2 and H2 will always react to form NH3 in this 1:3:2 ratio of moles.

  8. Stoichiometry • A BCE is essential for all calculations involving amounts of reactants and products. • A BCE is to stoichiometry as a recipe is to cooking.

  9. Mole-Mole Stoichiometry • The following equation shows the synthesis of aluminum oxide from its elements: • 4Al(s) + 3O2(g)  2Al2O3(s) • If you only had 1.8 moles of Al, how much product could you make? • Given: 1.8 mol Al • Want: ??? mol Al2O3 • Conversion factor: 4 mol Al produce 2 mol Al2O3 • 1.8 mol Al x 2 mol Al2O3 4 mol Al = 0.90 mol Al2O3

  10. Mole-Mole Stoichiometry • The following equation shows the synthesis of aluminum oxide from its elements: • 4Al(s) + 3O2(g)  2Al2O3(s) • If you wanted to produce 24 moles of Al2O3, how many moles of each reactant would you need? • Given: 24 mol Al2O3 • Want: ??? mol Al • Want: ??? mol O2 • Conversion factors: • 4 mol Al produce 2 mol Al2O3 • 3 mol O2 produce 2 mol Al2O3

  11. Mole-Mole Stoichiometry • If you wanted to produce 24 moles of Al2O3, how many moles of each reactant would you need? • 4Al(s) + 3O2(g)  2Al2O3(s) • 24 mol Al2O3 x • 24 mol Al2O3 x = 48 mol Al 4 mol Al 3 mol O2 2 mol Al2O3 2 mol Al2O3 = 36 mol O2

  12. moles of substance A moles of substance A BCE Mole-Mole Stoichiometry • You can always do mole-mole conversions if you have a BALANCED chemical equation. • Unbalanced equations are not useful for stoichiometry.

  13. Mole-Mole Stoichiometry Practice • Octane combusts in the presence of oxygen via the following reaction: • 2C8H18 + 25O2 16CO2 + 18H2O • How many moles of oxygen gas can react with 7.50 moles of octane? • 7.50 mol C8H18 x • If 48 moles of CO2 are produced during one such reaction, how many moles of H2O are also produced? • 48 mol CO2 x 25 mol O2 18 mol H2O = 93.8 mol O2 2 mol C8H18 16 mol CO2 = 54 mol H2O

  14. Mass-Mass Stoichiometry • No lab balance measures moles directly. • Mass-mass conversions are more common and practical in real life. • From the mass of one reactant or product you can calculate the mass of any other reactant or product in the same reaction. • Requires a balanced chemical equation. • As in mole-mole calculations, the unknown can be either a reactant or a product.

  15. Mass-Mass Stoichiometry • Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2): • CaC2 + 2H2O  Ca(OH)2 + C2H2 • How many grams of C2H2 are produced by adding water to 5.00 grams of CaC2? • Given: 5.00 g CaC2 • Want: ??? g C2H2 • Conversion factors: • 1 mol CaC2 produces 1 mol C2H2 • 1 mol CaC2 = 64.099 g CaC2 (molar mass of CaC2) • 1 mol C2H2 = 26.038 g C2H2 (molar mass of C2H2)

  16. Mass-Mass Stoichiometry • Make a plan! • Your first step should always be to change to moles! • Next, do a mole-mole conversion using the BCE to guide you. • Finally, convert from moles back to grams. • 5.00 g CaC2 mol CaC2  mol C2H2  g C2H2 • 5.00 g CaC2 26.038 g C2H2 1 mol C2H2 1 mol CaC2 x x = 2.03 g C2H2 x 64.099 g CaC2 1 mol C2H2 1 mol CaC2

  17. Mass-Mass Stoichiometry mass A moles A moles B mass B MMA BCE MMB

  18. Mass-Mass Stoichiometry • Copper metal reacts with concentrated nitric acid to produce copper(II) nitrate, nitrogen dioxide gas, and water: • Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O • How many grams of HNO3 are needed to react completely with 0.450 grams of Cu? • Given: 0.450 g Cu • Want: ??? g HNO3 • Conversion factors: • 1 mol Cu reacts with 4 mol HNO3 • 1 mol Cu = 63.546 g Cu (molar mass of Cu) • 1 mol HNO3 = 63.012 g HNO3 (molar mass of HNO3)

  19. Mass-Mass Stoichiometry • How many grams of HNO3 are needed to react completely with 0.450 grams of Cu? • Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O • Make a plan! • 0.450 g Cu  mol Cu  mol HNO3  g HNO3 • 0.450 g Cu 1 mol Cu 4 mol HNO3 63.012 g HNO3 = 1.78 g HNO3 x x x 63.546 g Cu 1 mol Cu 1 mol HNO3

  20. A Few Reminders • Stoichiometry problems can involve reactants, products, or both. • They can also involve moles, grams, liters or milliliters, atoms, molecules, or any other unit that measures the amount of a particular chemical. • All stoichiometry problems have a mole-mole conversion at their center. • At the very least, you’ll need a balanced chemical equation in order to solve any stoichiometry problem. • It’s a good idea to write not only the unit (g, mol, L, etc) but also the chemical formula when doing conversions. • Ex: write 48.0 g Cu(NO3)2, not just 48.0 g.

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