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Announcements

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.

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Announcements

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  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 5—if want preview look at sections 5.1&5.2.

  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 --> 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 --> 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) ––> 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 + … ––> bB + …: • g A –(÷MA)–> mol A –(•[b mol B/a mol A])–> mol B –(xMB)–> 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 ––> aA + bB.

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