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che106_lecture8

# che106_lecture8 - Chemistry 106 Lecture 8 Topics ...

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Unformatted text preview: Chemistry 106 Lecture 8 Topics: Limi6ng Reactants, Yields Chapter 3.6 - 3.7 EXAM 1 •  Exam #1 is this Wednesday, September 28th at regular class 6me (5.15- 6:35). •  The Exam will be held in here as for normal lecture. •  All three TA’s will be present •  Bring a calculator (scien6ﬁc is ﬁne but no graphing). •  No phones or laptops to be on or visible •  Exam is mul6ple- choice •  A prac6ce test is on blackboard Exam Details •  Exam #1 will cover Chapters 1 through 3. •  Primarily based on homework and lecture material. •  Exam will be mul6ple choice (30 ques6ons). •  You need to bring a #2 pencil. •  You need a calculator (not a cell phone). •  You will be given a copy of the Periodic Table from the textbook. Stoichiometry: Quan6ta6ve Rela6ons in Chemical Reac6ons •  Stoichiometry is the calcula6on of the quan66es of reactants and products involved in a chemical reac6on. –  It is based on the balanced chemical equa6on and on the rela6onship between mass and moles. –  Such calcula6ons are fundamental to most quan6ta6ve work in chemistry. Stoichiometric Calcula6ons The coeﬃcients in the balanced equa6on give the ra6o of moles of reactants and products. Stoichiometric Calcula6ons Based on the high- temperature reac6on between CH4 and H2O shown here, how many moles of each product can be obtained star6ng with 4.0 mol CH4? 2 H2O and 2 CH4 Ra6os 2 CH4 : 2 CO or 1:1 2 CH4 : 6 H2 or 1:3 2 CO and 6 H2 4.0 moles of CH4 would yield 4.0 moles of CO and 12.0 moles of H2 Molar Interpreta6on of a Chemical Equa6on •  The balanced chemical equa6on can be interpreted in numbers of molecules, but generally chemists interpret equa6ons as “mole- to- mole” rela6onships. –  For example, the Haber (or Haber- Bosch) process for producing ammonia involves the reac6on of hydrogen and nitrogen. N 2 (g ) + 3 H 2 (g ) → 2 NH 3 (g ) Molar Interpreta6on of a Chemical Equa6on •  This balanced chemical equa6on shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. N 2 (g) + 3H 2 (g) → 2 NH 3 (g ) 1 molecule N2 + 3 molecules H2 2 molecules NH3 1 mol N 2 + 3 mol H 2 → 2 mol NH 3 Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. Molar Interpreta6on of a Chemical Equa6on •  Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2. N 2 (g) + 3H 2 (g) → 2 NH 3 (g ) Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple. 2 mol NH 3 4.8 mol H 2 × = 3.2 mol NH 3 3 mol H 2 Mass Rela6onships in Chemical Equa6ons •  How many grams of HCl are required to react with 5.00 grams of manganese dioxide according to this equa6on? 4 HCl(aq) + MnO 2 (s ) → 2 H 2O(l) + MnCl 2 (aq) + Cl 2 (g ) Mass Rela6onships in Chemical Equa6ons 4 HCl(aq) + MnO 2 (s ) → 2 H 2O(l) + MnCl 2 (aq) + Cl 2 (g ) •  First, you write what is given (5.00 g MnO2) and convert this to moles. •  Then convert to moles of what is desired (mol HCl). •  Finally, you convert this to mass (g HCl). 36.5 g HCl 1 mol MnO 2 4 mol HCl × 5.00 g MnO 2 × × 86.9g MnO 2 1 mol MnO 2 1 mol HCl = 8.40 g HCl Limi6ng Reagent •  The limiBng reactant (or limiBng reagent) is the reactant that is en6rely consumed when the reac6on goes to comple6on. •  The limi6ng reagent ul6mately determines how much product can be obtained. •  For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made. limiting reactant Molecular View of 2H2 + O2 → 2H2O Reac6on with H2 as the Limi6ng Reactant Limi6ng Reagent •  Zinc metal reacts with hydrochloric acid by the following reac6on. Zn(s) + 2 HCl(aq) → ZnCl 2 (aq) + H 2 (g ) –  If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? Limi6ng Reagent •  Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiBng reagent. 1 mol H 2 0.30 mol Zn × = 0.30 mol H 2 1 mol Zn 1 mol H 2 0.52 mol HCl × = 0.26 mol H 2 2 mol HCl HCl must be limiting. •  Since HCl is the limiting reagent, the amount of H2 produced must be 0.26 mol. How many moles of O2 are required to react completely with 1.0 mole NO? 2 NO (g) + O2 (g) 2 NO2 (g) A.  0.5 mol O2 B.  1.0 mol O2 C.  1.5 mol O2 D. 2.0 mol O2 E.  2.5 mol O2 How many moles of O2 are required to react completely with 1.0 mole NO? 2 NO (g) + O2 (g) 2 NO2 (g) A. 0.5 mol O2 B.  1.0 mol O2 C.  1.5 mol O2 D. 2.0 mol O2 E.  2.5 mol O2 Theore6cal and Percent Yield •  The theoreBcal yield of product is the maximum amount of product that can be obtained from given amounts of reactants. –  The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). actual yield %Yield = × 100% theoretical yield Theore6cal and Percent Yield •  To illustrate the calcula6on of percentage yield, recall that the theore6cal yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2. •  If the actual yield of the reac6on had been 0.22 g H2, then 0.22 g H 2 %Yield = × 100% = 42% 0.52 g H 2 Next Lecture (ajer exam) •  Topics: Ions/Precipita6on and Acid- Base •  Text Reading: 4.1 - 4.3 ...
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