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Stoichiometry

Stoichiometry. Applying equations to calculating quantities . Learning objectives. Use chemical equations to predict quantity of one substance from given quantity of another Determine percentage yield Identify limiting reactant. The chemical equation. a A + b B = c C + d D

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Stoichiometry

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  1. Stoichiometry Applying equations to calculating quantities

  2. Learning objectives • Use chemical equations to predict quantity of one substance from given quantity of another • Determine percentage yield • Identify limiting reactant

  3. The chemical equation aA + bB = cC + dD The Law of Conservation of Matter states that matter is neither created nor destroyed All atoms on left must be same as those on right Reactant side coefficient ELEMENT or COMPOUND Product side

  4. Working with equations:STOICHIOMETRY • Predict how much product is obtained from given amount of reactant • Predict how much reactant is needed to give required amount of product • Predict how much of one reactant is required to give optimum result with given amount of another reactant

  5. Stoichiometry with equations:The roadmap • Equations are in moles, but we measure in grams • Three conversions required: • A is given substance; B is target substance • Must convert grams A to moles A using molar mass • Use coefficients in equation to get moles B from moles A • Convert moles B to grams B using molar mass

  6. Mole:mole ratio: the central step • Tells us molar ratio of substances in balanced chemical equation aA + bB = cC + dD • Mole:mole ratio B:A = b/a

  7. Types of problems:Moles A → moles B(A is given; B is target) aA + bB = cC + dD • Single step • Require mole:mole ratio: • Always target/given (B/A) • In balanced equation a mol A ≡ b mol B

  8. Moles A → mass B aA + bB = cC + dD Two steps on road map • Convert moles A → moles B: Mole:mole ratio (target/given): • Convert moles B → mass B using molar mass B

  9. Mass A → mass B aA + bB = cC + dD All three steps • Mass A → moles A using molar mass A • Moles A → moles B using mole:mole ratio • Moles B → mass B using molar mass B

  10. Summary of stoichiometry problems • Maximum of three conversions required • Must convert grams A to moles A using molar mass • Use coefficients in equation to get moles B from moles A • Convert moles B to grams B using molar mass • Maximum of three pieces of information required • Molar mass of given substance (maybe) • Molar mass of target substance (maybe) • Balanced chemical equation (always)

  11. Work this example • CH4 + 2O2 = CO2 + 2H2O • What mass of CO2 is produced by the complete combustion of 16 g of CH4 • Atomic weight H = 1, C = 12, O = 16 44 g • Do stoichiometry exercises

  12. Reaction Yield • The actual yield from a chemical reaction is normally less than predicted from the stoichiometry. • Incomplete reaction • Product lost in recovery • Competing side reactions • Percent yield is:

  13. Worked example • Actual yield of product is 32.8 g after reaction of 26.3 g of C4H8 with excess CH3OH to give C5H12O. • What is theoretical yield? Use stoichiometry to get mass of product: convert mass  moles  moles  mass • Theoretical yield = 41.4 g • Percent yield = 32.8/41.4 x 100 %

  14. Limiting reactant • Balanced equations involve precisely the right amounts of each reactant. None is left over • Real-life situations usually involve an excess of one reactant • Burning CH4 in excess O2 • Reacting Zn with excess HCl

  15. Identifying the limiting reactant • In balanced equations each reactant will give the same amount of products • When one reactant is limiting (deficient) the amount of product will be determined by its amount

  16. Identifying the limiting reactant • Which reactant gives the least amount of product CO(g) + 2 H2(g) = CH3OH(g) • Given 3 mol CO and 5 mol H2 • 3 mol CO → 3 mol CH3OH • (mol:mol ratio CH3OH:CO = 1) • 5 mol H2→ 2.5 mol CH3OH • (mol:mol ratio CH3OH:H2 = 0.5) • H2 is limiting • Maximum product = 2.5 mol CH3OH

  17. Solution stoichiometry • In solids, moles are obtained by dividing mass by the molar mass • In liquids, it is necessary to convert volume into moles using the concentration

  18. Solution stoichiometry • How much volume of one solution to react with another solution • Given volume of A with molarity MA • Determine moles A • Determine moles B • Find volume of B with molarity MB

  19. Titration • Use a solution of known concentration to determine concentration of an unknown • Must be able to identify endpoint of titration to know stoichiometry • Most common applications with acids and bases

  20. Example • How much 0.125 M NaHCO3 is required to neutralize 18.0 mL of 0.100 M HCl?

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