Announcements - PowerPoint PPT Presentation

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Announcements

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... MM of ... MM(CaO) =56.077 g/mol. MM(CO2) = 44.010 g/mol. MM(CaCO3) ... MM(Ca(OH)2) = 74.0926 g/mol. CaCO3 -- CaO CO2. Example Limiting Reagent ... – PowerPoint PPT presentation

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Title: Announcements


1
Announcements
  • To join clicker to class today
  • Turn on the Clicker (the red LED comes on).
  • Push Join button followed by 20 followed by
    the Send button (switches to flashing green LED
    if successful).
  • Exam next Wed.
  • Quiz on sections 4.2 4.8 Monday.

  • Monday discussion review.
  • Monday lecture may have intro to chapter 5if
    want preview look at sections 5.15.2.

2
Review
  • Calculating MM of compounds.
  • Stoichiometry used to figure out masses or moles
    of reagents used or products produced in a
    chemical reaction.
  • Calculating yield.
  • Finished with clicker question

3
Final Clicker Question
  • Heat 200.00 g CaCO3 (limestone) to get CaO CO2,
    get 80.00 g CO2. What is yield of CO2 (or of
    CaCO3 decomposed)? You were given molar masses
    to save time.
  • 85 got the right answer 90.97
  • The most common wrong answer was 40.
  • Path to correct answer
  • Balance equation CaCO3 --gt CaO CO2
  • Calculate g of CO2 expected from 200.00 g of
    CaCO3
  • Calculate yield
  • 100(80.00 g CO2/expected g CO2)

4
Other possible questions with same info.
CaCO3 --gt CaO CO2
  • How many g of CaO do I get when 80.00 g of CO2
    are produced?
  • MM(CaO) 56.077 g/mol
  • MM(CO2) 44.010 g/mol
  • MM(CaCO3) 100.087 g/mol
  • Concrete is made by slaking CaO with water to
    make Ca(OH)2.How much Ca(OH)2 will I get when the
    CaO is slaked?
  • MM(Ca(OH)2) 74.0926 g/mol

5
Example Limiting Reagent Calculation
  • RXN 3Na2CO3(aq) 2H3PO4(aq) gt 2Na3PO4(aq)
    3CO2(g) 3H2O(l)
  • Have 5.00 g of each reagent. How much CO2 do we
    get?
  • M(Na2CO3) 105.988 g/mol Na2CO3
  • M(H3PO4)97.994 g/mol H3PO4
  • M(CO2)44.010 g/mol CO2

6
Composition useful data for examples
  • M(H2O) 18.0153 g H2O/mole H2O.
  • Molecules called ethylene and propylene are both
    14.372 by mass H. The rest is carbon (85.628).
  • M(ethylene) 28.054 g/mol
  • M(propylene) 42.081 g/mol

7
Review
  • Stoichiometry used to figure out masses of
    reagents used or products produced in a chemical
    reaction.
  • Percent yield
  • Reviewed limiting reagents.
  • After balancing chemical equations all of these
    calculations require doing the following series
    of conversion given aA gt bB
  • g A (MA)gt mol A (b mol B/a mol A)gt mol B
    (xMB)gt g B.
  • For limiting reagent we do this process up to the
    moles of B stage for all reagents and only do
    the last step for the smallest moles of B.
  • Note that sometimes A and B will be on the same
    side of the balanced chemical equation cC dD
    gt aA bB.
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