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Stoichiometry

Stoichiometry. Stoichiometry. Is the study of the relative quantities of reactants and products in chemical reactions  the mole ratio. How can you get to the moon?. You need ROCKET FUEL!. 2N 2 H 4 (l) + N 2 O 4 (l)  3N 2 (g) + 4H 2 O(g) + ENERGY

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Stoichiometry

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  1. Stoichiometry

  2. Stoichiometry • Is the study of the relative quantities of reactants and products in chemical reactions •  the mole ratio

  3. How can you get to the moon? You need ROCKET FUEL! 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY (hydrazine) (dinitrogen tetroxide) If 1000 tonnes of N2O4 is to be used, how much N2H4 should be used?

  4. Ratios in Balanced Chemical Equations 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY This equation means: 2 molecules of N2H4 react with 1 molecules of N2O4 • 200 molecules of N2H4 react with ______ molecules of N2O4 100 • 4 molecules of N2H4 react with ______ molecules of N2O4 2 8 million • 16 million molecules of N2H4 react with _________ molecules of N2O4 OR 2 moles of N2H4 react with 1 mole of N2O4

  5. The reactants and products in a balance chemical equation will always be related by ratios established by the coefficients in the equation.

  6. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY a.) How many moles of N2H4 are required to react with 8 moles of N2O4? Step 1: Check the equation is balanced 2 mols of N2H4 1 mols of N2O4 Step 2: Set up mole ratio Step 3: set equal to given and unknown 2 mols of N2H4 = x mols of N2H4 1 mols of N2O4 8 mols of N2O4 Cross multiply! x mols of N2H4 X 1 mols of N2O4= 2 mols of N2H4 X 8 mols ofN2O4 x mols of N2H4 = 2 mols of N2H4 X 8 mols ofN2O4 1 mols of N2O4 x mols of N2H4 = 16 mols of N2H4

  7. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY b.) How many moles of N2H4 are required to react with 17 moles of N2O4? x mols of N2H4 = 2 mols of N2H4 17 mols of N2O4 1 mols of N2O4 x mols of N2H4= 2 mols of N2H4 X 17 mols of N2O4 1 mols of N2O4 = 34 mols of N2H4

  8. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY c.) How many moles of N2H4 are required to react with 0.23 moles of N2O4? x mols of N2H4 = 2 mols of N2H4 0.23 mols of N2O4 1 mols of N2O4 x mols of N2H4= 2 mols of N2H4 X 0.23 mols of N2O4 1 mols of N2O4 = 0.46 mols of N2H4

  9. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY d.) How many moles of H2Oare produced from 8 moles of N2H4? x mols of H2O= 4 mols of H2O 8 mols of N2H42 mols of N2H4 x mols of H2O= 4 mols of H2O X 8 mols of N2H4 2 mols of N2H4 = 16 mols of H2O

  10. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY e.) How many moles of N2 are produced from 11 moles of N2O4? x mols of N2 = 3 mols of N2 11 mols of N2O41 mols of N2O4 x mols of N2= 3 mols of N2 X 11 mols of N2O4 1 mols of N2O4 = 33 mols of N2

  11. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY f.) If 0.55 moles of H2Oare produced, How much N2H4 reacted? x mols of N2H4= 2 mols of N2H4 0.55 mols of H2O4 mols of H2O x mols of N2H4= 2 mols of N2H4 X 0.55 mols of H2O 4 mols of H2O = 0.28 mols of N2H4

  12. 2N2H4 (l) + N2O4 (l)  3N2(g) + 4H2O(g) + ENERGY g.) If 0.55 moles of H2Oare produced, How much N2 is produced? x mols of N2 = 3 mols of N2 0.55 mols of H2O4 mols of H2O x mols of N2= 3 mols of N2 X 0.55 mols of H2O 4 mols of H2O = 0.41 mols of N2

  13. Phosphorus can be combined with oxygen gas to form diphosphorus trioxide. How many molecules of diphosphorus trioxide can be made from 5 moles of phosphorus? • Write out balanced chemical equation 4P(s) + 3O2(g)  2P2O3(s) x mols of P2O3 = 2 mols of P2O3 5 mols of P4 mols of P x mols of P2O3= 2 mols of P2O3 X 5 mols of P 4 mols of P = 2.5 mols of P2O3 Molecules of P2O3 = 2.5 mols of P2O3X 6.022 X 1023 molecules/mols of P2O3 = 1.5 x 1024 molecules of P2O3

  14. Assignment : Practice Problems pages 237-238 #1-7, page 240 #9, 10

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