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Stoichiometry – Chemical Quantities Notes

Stoichiometry – Chemical Quantities Notes. Stoichiometry. Stoichiometry – Study of quantitative relationships that can be derived from chemical formulas and chemical equations Mole ─ Mole Relationship need a balanced equation

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Stoichiometry – Chemical Quantities Notes

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  1. Stoichiometry – Chemical Quantities Notes

  2. Stoichiometry Stoichiometry – Study of quantitative relationships that can be derived from chemical formulas and chemical equations Mole ─ Mole Relationship need a balanced equation Mole Ratio – the ratio of moles of one substance to moles of another substance in a balanced chemical equation The coefficients in a balanced equation give the relative numbers of molecules, as well as, the relative number of moles. CO(g) + 2H2(g) CH3OH(l) 1 mol CO = 2 mol H2 = 1 mol CH3OH Ex: How many moles of O2 are required to produce 10. moles of CO2? 2 CO + O2 2 CO2 10. mol CO2 1 mol O2 x __________ 5.0 mol O2 = 2 mol CO2

  3. What other relationships do we have for the mole? • 1 mol = 6.022 x 1023 atoms / molecules / particles • 1 mol = [molar mass] g We can add these mole relationships on either end of the mole ratio: # unit A x 1 mol A x mol B x __ unit B = # unit B _ unit A _ mol A 1 mol B mole relationship mole ratio mole relationship (switch units) (switch substances) (switch units) *A is the GIVEN substance & B is the WANTED substance

  4. K 1 x 39.1 = 39.1 Cl 1 x 35.5 = 35.5 O 3 x 16.0 = 48.0 122.6 g/mol Mass A – Mole B Ex: Calculate moles of O2 produced if 2.50 g KClO3 decomposes completely: 2 KClO3 2 KCl + 3 O2 2.50 g KClO3 1 mol KClO3 3 mol O2 x ________________ x ______________ = 122.6 g KClO3 2 mol KClO3 0.0306 mol O2

  5. Na 1 x 23.0 = 23.0 Cl 1 x 35.5 = 35.5 58.5 g/mol Mass A – Mass B Ex: Determine the mass of NaCl that decomposes to yield 355 g Cl2 2NaCl  2 Na + Cl2 1 mol Cl2 2 mol NaCl 58.5 g NaCl 355 g Cl2 x _____________ x ______________ x ______________ = 71.0 g Cl2 1 mol Cl2 1 mol NaCl Cl 2 x 35.5 = 71.0 g/mol 585 g NaCl

  6. Mole A – Mass B Ex: Calculate the number of grams of oxygen required to react exactly with 4.30 mol of propane, C3H8, in the reaction by the following balanced equation: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 5 mol O2 32.0 g O2 4.30 mol C3H8 x _____________ x __________ = 1 1 mol O2 mol C3H8 688 g O2 O 2 x 16.0 = 32.0 g/mol

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