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Stoichiometry 2

Stoichiometry 2. Mole-Mass, Mass-Mole, and Other Conversions. More Stoichiometry!. In the last lesson we learned how to do mole-mole conversions and mass-mass conversions. We need a balanced chemical equation (BCE) for all stoichiometric conversions.

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Stoichiometry 2

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  1. Stoichiometry 2 Mole-Mass, Mass-Mole, and Other Conversions

  2. More Stoichiometry! • In the last lesson we learned how to do mole-mole conversions and mass-mass conversions. • We need a balanced chemical equation (BCE) for all stoichiometric conversions. • We need molar masses for mass-mole and mole-mass conversions.

  3. Stoichiometry 2 • It would be nice if we only ever had to solve mole-mole and mass-mass problems, but that’s not the case. • We should be prepared to solve any kind of stiochiometry problem, given the right tools.

  4. Stoichiometry Map grams of A grams of b particles of b particles of A moles of B moles of A Liters of solution A Liters of solution b Liters of gas A Liters of gas b milliLiters of solution A milliLiters of solution b milliLiters of gas A milliLiters of gas b

  5. Stoichiometry 2 • HAsO2 decomposes when heated into diarsenic trioxide and water: • 2HAsO2 As2O3 + H2O • How many moles of water are produced by the decomposition of 40.0 g HAsO2? • Given: 40.0 g HAsO2 • Want: ??? mol H2O • Conversion factors: • 2 mol HAsO2 produce 1 mol H2O • 1 mol HAsO2 = 107.93 g/mol

  6. 1 mol HAsO2 1 mol H2O x x 107.93 g HAsO2 2 mol HAsO2 Stoichiometry 2 • HAsO2 decomposes when heated into diarsenic trioxide and water: • 2HAsO2 As2O3 + H2O • How many moles of water are produced by the decomposition of 40.0 g HAsO2? • Plan your solution! • 40.0 g HAsO2  mol HAsO2  mol H2O • 40.0 g HAsO2 = 0.185 mol H2O

  7. Stoichiometry 2 • Al2O3 + Na2CO3 2NaAlO2 + CO2 • What mass of Al2O3 is needed to form 2.50 moles of NaAlO2? • Given: 2.50 mol NaAlO2 • Want: ??? g Al2O3 • Conversion factors: • 1 mol Al2O3 produces 2 mol NaAlO2 • 1 mol Al2O3 = 101.96 g/mol • Plan your solution! • 2.50 mol NaAlO2 mol Al2O3  g Al2O3

  8. 1 mol Al2O3 101.96 g Al2O3 x x 2 mol NaAlO2 1 mol Al2O3 Stoichiometry 2 • 2.50 mol NaAlO2 = 127 g Al2O3

  9. Stoichiometry 2 • You can do any sort of stoichiometric conversion as long as you have the appropriate conversion factor!

  10. Stoichiometry 2 • How many molecules of CO2 are produced by the burning of 3.80 mg of C3H8? • C3H8 + 5O2 3CO2 + 4H2O • Given: 3.80 mg C3H8 • Want: molecules of CO2 • Conversion factors: • 1 mol C3H8 produces 3 mol CO2 • 1 mg = 0.001 g • 1 mol C3H8 = 44.10 g C3H8 • 1 mol = 6.022x1023 molecules • Plan your solution! • 3.80 mg C3H8 g C3H8  mol C3H8  mol CO2  molecules CO2

  11. Stoichiometry 2 3.80 mg C3H8 = 1.56x1020 molecules

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