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Chapter 3 (CIC) and Chapter 3, 18 (CTCS)

Chapter 3 (CIC) and Chapter 3, 18 (CTCS). Read in CTCS Chapter 3.4, 6, 7, and 18.4 (pgs. 697-699) Problems in CTCS: 3.29, 31, 33, 35, 39, 41, 59, 63, 69, 71, 73, 79, and 18.27. How Much CO 2 is 3 bmt of C?. Several ways to solve (%C in CO 2 and stoichiometric relationships)

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Chapter 3 (CIC) and Chapter 3, 18 (CTCS)

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  1. Chapter 3 (CIC) and Chapter 3, 18 (CTCS) • Read in CTCS Chapter 3.4, 6, 7, and 18.4 (pgs. 697-699) • Problems in CTCS: 3.29, 31, 33, 35, 39, 41, 59, 63, 69, 71, 73, 79, and 18.27

  2. How Much CO2 is 3 bmt of C? • Several ways to solve (%C in CO2 and stoichiometric relationships) Q: What is the %C in CO2? A: 27.3% • This is equivalent to dividing 3 by 0.273 (27.3%)

  3. Stoichiometric Relationships • C + O2 CO2 • Since a bmt is so much larger than an amu, it is helpful to work on a larger scale than an amu – such as grams • If a 12C atom weighs 12.0000…amu by definition, how many 12C atoms does it take to weigh 12.000…g? 6.02 x 1023 (Avogadro’s number) • Dozen, Gross, Ream, Mole

  4. Inverted: 6.02 x 1023 amu/g • How big is Avogadro’s number? • 1 mol of seconds = 4 million times as long as Earth has existed • We use Avogadro’s number to count individual atoms, molecules, electrons, etc.

  5. For MgCO3 we were finding ~ 0.40 g CO2 released. Had we not known it was a carbonate, could we have figured it out knowing that 0.73 g of HCl was used in the reaction? CO32- + 2 HCl  H2O + CO2 + 2 Cl- HCO3- + HCl  H2O + CO2 + Cl- How many g of that 0.40 g is due to C (27.3%)? Stoichiometry Lab

  6. How do you know it was Mg? • Suppose the slope of the line in the 1st week’s data had been 0.5g CO2/1 g XCO3. • ? = 88 g XCO3 and of that 88g, 44 + 16 (or 60g) of it was due to the CO3. This means that 28 g of it was due to X. If X was an alkaline earth metal then it must mass 28g. If X was an alkali metal X must mass 14 g because the formula is X2CO3.

  7. Q: If volcanoes release 19 mmt SO2, how much S is “combusted”? A: 9.5 mmt S Q: If a lighter holds 5.0g of butane (C4H10), how many grams of CO2 are produced upon combustion? A: 15 g CO2 g (reactant)  mol (reactant)  mol (product)  g (product)

  8. Limiting Reagent CO32- + 2 H+ H2O + CO2 • When adding small amounts of the carbonate a pink solution remained. • What was the limiting reagent? • What was the excess (xs) reagent? • Could you tell by looking at the solution? • When large amounts of carbonate were added, a white solid and colorless solution appeared. • What was the limiting reagent? • What was the excess (xs) reagent? • Could you tell by looking at the solution? • At what point on the graph could you use the stoichiometric relationships?

  9. Percent Yield (actual/theoretical x 100) • In reality, we ended up getting about 0.35 g CO2 from the MgCO3 instead of the 0.44 g that is the theoretical amount. See next slide for the graph! Q: What is the percent yield of CO2 that was obtained? A: 80% • What does this mean? • HCl concentration was not 0.73 g • A poor yielding reaction is not suitable for this lab

  10. Other Gases and What We Know • CH4 is 30x as effective as a greenhouse gas as CO2 but its’ concentration is only at 1.7 ppm • CFC’s – 25,000x effective and [CFC] is at 0.28 ppb • Modeling earth’s Temp is still crude • We know • CO2 increases global Temp • [CO2] increased over the last century • [CO2] increased due to human activity • Global Temp increased over last century • Uncertain about • CO2 and other gases are responsible for increased Temp • Avg global Temp will continue to increase as anthropogenic greenhouse gases increase

  11. What Should We Do? • Act now! Is the cure worse than the disease? • Fertilize ocean with iron to increase phytoplankton • Release S containing compounds into air • Pump CO2 to bottom of ocean • Pollution Rights • US has a great head-start over undeveloped countries • Study the problem more • Nothing – It’s inevitable - Adapt • Fossil fuels are too valuable to just combust

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