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Stoichiometry II Dr. Ron Rusay Spring 2008 © Copyright 2008 R.J. Rusay.

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Presentation on theme: "Stoichiometry II Dr. Ron Rusay Spring 2008 © Copyright 2008 R.J. Rusay."— Presentation transcript:

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2 Stoichiometry II Dr. Ron Rusay Spring 2008 © Copyright 2008 R.J. Rusay

3 Chemical Equations / Stoichiometric Calculations Example:Example: C 8 H 18(l) + O 2(g) -----> CO 2(g) + H 2 O (l) For the combustion of octane which producesFor the combustion of octane which produces carbon dioxide and water. The equation must conserve mass,i.e. accountThe equation must conserve mass,i.e. account for the mass in product and reactant by having a “balance” which is an equal number of individual atoms in reactant and product. © Copyright R.J. Rusay

4 Chemical Equations  The equation is “balanced” as before by placing stoichiometric factors before each of the molecules in the reaction so that the atoms equal ? C 8 H 18(l) + ? O 2(g) -----> ? CO 2(g) + ? H 2 O (l) 1 25/2 89 Remove fraction: 2 C 8 H 18(l) + 25 O 2(g) -----> 16 CO 2(g) +18 H 2 O (l) © Copyright R.J. Rusay

5 Mass Calculations  The Balanced Equation is a balance on a molar basis which can be related to mass. 2 C 8 H 18(l) + 25 O 2(g) -----> 16 CO 2(g) +18 H 2 O (l) 228 g of octane (2 moles)* will react with 800 g of oxygen (25 moles) to produce 704 g of carbon dioxide and 324 g of water. *(2 moles octane x 114 g/mol = 228 g ) © Copyright R.J. Rusay

6 1.Balance the chemical equation. 2.Convert mass of reactant or product to moles. 3.Identify mole ratios in balanced equation: They serve as the “Gatekeeper”. 4.Calculate moles of desired product or reactant. 5.Convert moles to grams. Mass Calculations: Products Reactants © Copyright R.J. Rusay

7 Mass Calculations: Products Reactants © Copyright R.J. Rusay

8 Mass Calculations: Products Reactants © Copyright R.J. Rusay

9 Mass Calculations: Reactants Products � How many grams of salicylic acid are needed to produce 1.80 kg of aspirin? � Balanced Equation:

10 Mass Calculations: Reactants Products © Copyright R.J. Rusay grams (Aspirin) grams (Salicylic Acid) 1800 grams (A) grams (A) (Molecular Weight A) 1 mol (A) grams (SA) ? (SA) 1 mol (SA) (Molecular Weight SA) Avogadro's Number Atoms Molecules Stoichiometry 1 mol SA 1 mol A "Gatekeeper" Calculation:

11 Mass Calculations: How many grams of salicylic acid are needed to produce 1.80 kg of aspirin? � Balanced Equation: ?g C7 = 1.80 x 10 3 g C9 x [mol C9 /180.15g C9 ] x [1 mol C7 / 1 mol C9 ] x g C7 /mol C7 1 1 ?g C7 = 1380 grams

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13 Limiting Reagent An Ice Cream Sundae ??????

14 Limiting Reagent An Ice Cream Sundae What’s left? What’s totally consumed?

15 Limiting Reagent

16 QUESTION

17 ANSWER ) 5)none of these Section 3.10 : Calculations Involving a Limiting Reactant(p. 106 The limiting reactant in a reaction is the reactant that produces the least number of grams of any product.

18 Mass Applications: Limiting Reagent  How do masses of reactants relate? Is there enough mass of each reactant for the reaction to consume all of both of them or will there be some left of one of them? 2 C 8 H 18(l) + 25 O 2(g) -----> 16 CO 2(g) +18 H 2 O (l) What would happen if only 600. g of O 2 were available for the reaction of 228 g of octane? © Copyright R.J. Rusay

19 Mass Applications: Determining a Limiting Reagent � Does one of the reactants have fewer stoichiometrically adjusted moles than the other reactant? If so, the reactant with the smaller value is the limiting reagent. Calculation: � Divide the mass of each reactant by its respective Molar Mass and by its Stoichiometric factor from the balanced equation; then compare the results.

20 Limiting Reagent Calculation  The reactant present in the smallest molar amount considering stoichiometry limits the mass basis of any reaction. 2 C 8 H 18(l) + 25 O 2(g) -----> 16 CO 2(g) +18 H 2 O (l) 228 g octane / 114 g/mol = 2 mol octane 600. g oxygen / 32 g/mol = mol oxygen 2 mol octane / 2 mol (stoich.) = mole oxygen / 25 mol (stoich.) = 0.75 © Copyright R.J. Rusay

21 Mass Effects of the Limiting Reagent What amount of octane remains unreacted in the reaction of 600. g of O 2 with 228 g of octane?  600. g O 2 x mol O 2 /32g O 2 x [2mol C 8 H 18 /25mol O 2 ] x 114 g / mol C 8 H 18 = 171 g C 8 H 18 are reacted  228 g g = 57 g C 8 H 18 remain unreacted © Copyright R.J. Rusay

22 Limiting Reagent / Theoretical Yield The limiting reagent governs the theoretical yield of products. For the reaction of 228 g of octane with 600. g of oxygen, what is the theoretical yield of carbon dioxide? 2 C 8 H 18(l) + 25 O 2(g) -----> 16 CO 2(g) +18 H 2 O (l) 600. g O 2 x mol O 2 /32g O 2 x 16mol CO 2 /25mol O 2 x 44g / mol CO 2 = 528 g CO 2 © Copyright R.J. Rusay

23 Thoughts to Consider How much CO 2 do you produce per gallon of gasoline (octane, d= 0.70 g/ml) when gasoline is combusted? How much CO 2 do you personally produce from driving every week?…. every month? …. every year? ….from other uses and sources? Why do people in developed nations, like the U.S., Japan & in the EU, produce tons more of CO 2 per person than people in under developed or developing nations? Does the increase in “man-made” CO 2 relate to global warming?

24 Greenhouse Gases: The Chemistry of Warming

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26 What is a greenhouse gas? The sun’s energy & the molecule’s shape decide. Our atmosphere (air) is 78% nitrogen and 21% oxygen.Our atmosphere (air) is 78% nitrogen and 21% oxygen. Neither are greenhouse gases. They do not absorb infrared radiation (heat).Neither are greenhouse gases. They do not absorb infrared radiation (heat). However, H 2 O and CO 2 can absorb infrared energy. Without them earth would be very chilly.However, H 2 O and CO 2 can absorb infrared energy. Without them earth would be very chilly.

27 QUESTION

28 ANSWER ) )16.4 g Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products(p. 102 The balanced equation is Ca(OH) + 2HNO  Ca(NO) + 2HO. Water is the other product for an acid/base reaction.

29 Percent Yield  In synthesis, the actual yield (g) is measured and compared to the theoretical yield (g). This is the percent yield: % = actual / theoretical x 100  If a reaction produced 2.45g of Ibogaine, C 20 H 26 N 2 O, a natural product with strong promise in treating heroin addiction, and the theoretical yield was 3.05g, what is the % yield? © Copyright R.J. Rusay % yield = 2.45g / 3.05g x 100 = 80.3%

30 QUESTION The amount of tungsten metal, used to make many lightbulb filaments, starting with 25.0 grams of WO 3 in the following reaction was 18.0 grams. What percent yield does this represent? WO H 2  W + 3 H 2 O % 2.110% % 4.I think I may have confused one of the steps in this calculation

31 ANSWER Choice 3 has properly converted the initial grams of WO 3 to moles and used the mole ratio of the equation to determine the maximum moles of W that would be produced if all the WO 3 reacted (theoretical yield – 19.8 g). The division of the actual yield by the theoretical yield  100 produces the answer. Section 3.8: Stoichiometric Calculations: Amounts of Reactants and Products

32 Percent Yield � A reaction was conducted that theoretically would produce moles of quinine, C 20 H 24 N 2 O 2. The actual amount of isolated quinine was 780 mg. What is the percent yield of quinine? � 324 g/mol x mol = 81g = 810mg(theoretical) � % Yield = 780 mg/ 810 mg x 100 � % Yield = 96%

33 Mass Calculations: How many grams of salicylic acid are needed to produce 1.80 kg of aspirin if the process produces an 85.0% yield? � Balanced Equation: ?g C7 = 1.80 x 10 3 g C9 x [mol C9 /180.15g C9 ] x [1 mol C7 / 1 mol C9 ] x g C7 /mol C7 1 1 ?g C7 = 1380 grams ?g C7 = 2.12 x 10 3 g C9 x [mol C9 /180.15g C9 ] x [1 mol C7 / 1 mol C9 ] x g C7 /mol C7 ?g C7 = 1624 grams 1.80 x 10 3 g C9 /85/100 = 2.12 x 10 3 g C9 = 1380 grams / 85/100


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