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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,

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Presentation on theme: "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,"— Presentation transcript:

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

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

3 Stoichiometric Relationships C + O 2  CO 2 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 12 C atom weighs …amu by definition, how many 12 C atoms does it take to weigh …g? 6.02 x (Avogadro’s number) Dozen, Gross, Ream, Mole

4 Inverted: 6.02 x 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 Stoichiometry Lab For MgCO 3 we were finding ~ 0.40 g CO 2 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? CO HCl  H 2 O + CO Cl - HCO HCl  H 2 O + CO 2 + Cl - How many g of that 0.40 g is due to C (27.3%)?

6 How do you know it was Mg? Suppose the slope of the line in the 1st week’s data had been 0.5g CO 2 /1 g XCO 3. ? = 88 g XCO 3 and of that 88g, (or 60g) of it was due to the CO 3. 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 X 2 CO 3.

7 Q: If volcanoes release 19 mmt SO 2, how much S is “combusted”? A: 9.5 mmt S Q: If a lighter holds 5.0g of butane (C 4 H 10 ), how many grams of CO 2 are produced upon combustion? A: 15 g CO 2 g (reactant)  mol (reactant)  mol (product)  g (product)

8 Limiting Reagent CO H +  H 2 O + CO 2 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 CO 2 from the MgCO 3 instead of the 0.44 g that is the theoretical amount. See next slide for the graph! Q: What is the percent yield of CO 2 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

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11 Other Gases and What We Know CH 4 is 30x as effective as a greenhouse gas as CO 2 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 –CO 2 increases global Temp –[CO 2 ] increased over the last century –[CO 2 ] increased due to human activity –Global Temp increased over last century Uncertain about –CO 2 and other gases are responsible for increased Temp –Avg global Temp will continue to increase as anthropogenic greenhouse gases increase

12 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 CO 2 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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