20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant. 8.1 g H2 , 2.9 x 102 g H2O 17. How many litres of O2 gas are required to produce 100. g Al2O3?
1 / [CO2] C. [CaO][CO2] / [CaCO3] D. [CaCO3] / [CaO][CO2] _____ 13. The value of Kp for the reaction 2 NO2 (g) [pic] N2O4 (g) is 1.52 at 319 K. What is the value of Kp at this temperature for the reaction N2O4 (g) [pic] 2 NO2 (g) ? A. -1.52 B. 1.23 C. 5.74 X 10-4 D. 0.658 _____ 14.
2-propanol (bp=82 degrees C) 3. tetrahydofuran (bp=65 degrees C) 4. 1-butanol (bp=118 degrees C) 5. butanone (bp=80 degrees C) Give a better separation for the mixture to be distilled tetrahydofuran (bp=65 degrees C) because it is farthest from 100 degrees C Which alkyl halide would react fastest in a nucleophilic substitution using silver nitrate in ethanol (weak nucleophile, protic solvent)? 3-bromo-3-methylpentane (most
2) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.43/1.01) x100% = 42.57% That concludes that the percent recovery is 42,57%. 3) Percent recovery for isolation of hydroquinone dimethyl ether % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.16/1.01) x100% = 15.84% That concludes that the percent recovery is 15.84%. Table 2: : Experimental IR peaks compared to literature IR peaks for Benzoic acid Functional groups | Experimental peak (cm-1) | Literature peak (cm-1) | O-H | 3407-2563 | 3400-2564 | C=O | 1689 | 1689 | C-H |
a) for a we first need to find a balanced equation for when the hydrocarbons combust to form CO2 and H20. Then we plug in the deltaHf values and plug these into the equation. a) C4H6 + 11/2O2 ==> 4CO2 + 3H2O Delta Hrxn = [4DeltaHf(CO2)+3DeltaHf(H2O)] - [DeltaHf(C4H6) + 11/2DeltaHf(O2)] = [4(-393.5kJ) + 3(-285.83kJ)] - [111.9kJ + 11/2(0kJ)] = -2543.39kJ C4H8 + 6O2 ==> 4CO2 + 4H2O Delta H rxn = [4DeltaHf(CO2) + 4DeltaHf(H2O)] - [DeltaHf(C4H8) + 6DeltaHf(O2)] = [4(-393.5kJ) + 4(-285.83kJ)] - [1.2kJ +6(0kJ)] = -2718.52kJ C4H10 +13/2O2 ==> 4CO2 +5H2O DeltaHrxn = [4DeltaHf(CO2) + 5DeltaHf(H2O)] - [DeltaHf(C4H10) + 13/2DeltaHf(O2)] = [4(-393.5kJ) + 5(-285.83kJ)] - [-124.7kJ +
(2 marks) 8 Show 2 possible products that could form when Compound H undergoes a halogenation reaction with iodine. Draw the structural isomers and name them. 2 marks 9a Is this halogenation reaction is an addition or substitution reaction? ½ mark 9b Under what conditions would this reaction occur? ½ mark 10 Outline the reaction pathway to produce propanoic acid from propane.
| A) | CH3CH2O– | B) | CH3CH2O+ | C) | CH3CH2OH2+ | D) | CH3CH2OH3+ | 5. | Which one of the following mechanistically depicts the protonation of methanol by hydrogen bromide? | A) | A | B) | B | C) | C | D) | D | 6. | Give the molecular formula of the compound shown below: | A) | C8H16O | B) | C9H18O | C) | C10H18O | D) | C10H20O | 7. | The most stable resonance contributor of this would be: | A) | A | B) | B | C) | C | D) | D | 8.
RCH2G Primary R2CHG Secondary R3CG Tertiary. 4.7 Preparation of Alkyl Halides from Alcohols and Hydrogen Halides R-OH + H-X -> R-x +HOH “Alcohol” + “Hydrogen halide” -> “Alkyl halide” + “Water” The more primary the alcohol the more reactive, Primary=any halide Secondary=usually just Br with heat 4.8 Mechanism of the Reaction of Alcohols with Hydrogen Halides: Hammond’s Postulate Reaction of alcohol with hydrogen halide is a substitution Hydrogen Halide protenates Alcohol which then goes to H2O and the positive “alcohol” (no longer an alcohol since missing OH group) joins with the Halide Electrophile: electron loving Nucleophile: Nucleus seekers (extra electron pair) 4.9 Potential Energy Diagrams for Multistep Reactions: The SN1 Mechanism SN1 Mechanism: slow step is unimolecular 4.10 Structure, Bonding, and Stability of Carbocations Primary Carbocations do not exist in intermediates since too unstable (H3C)-C(H2) and (H3)-C are even more unstable. Secondary (H3C)2-C(H) are possible particularly with secondary alcohols with hydrogen
Synthesis of Methyl Stearate Post-Lab Submitted by Matthew Sharma TA: Evan Determining the Limiting Reagent Methyl Oleate (MW= 296.49 g/mol) Amount used = 1.141g (1.141g)(1mol / 296.49g) = 0.003848 mol ∴ The limiting reagent for the reaction is methyl oleate because hydrogen is used in excess Theoretical Yield Methyl Stearate (0.003848 mol)(298.504g/mol product) = 1.1486 g methyl stearate Percent Yield Given the 1:1 stoichiometry of the reaction: 100% x (0.2551g product / 1.1486g theoretical yield) = 22.2% Conclusion A 22.2% yield of synthesized methyl stearate was obtained via the catalytic hydrogenation of methyl oleate in the presence of the catalyst 10% palladium on carbon. The product
Title: Bromination of (E) – Stilbene (Microscale Procedure) Author’s Name: Reinaldo George Professor: Elvis Barrett Date of Experiment: Thursday July 16th 2015 Institution: Nova Southeastern University Abstract The purpose of this experiment was to synthesize the second intermediate in the b series of Sequential Reactions by carrying out the bromination of (E)-stilbene to obtain meso-stilbene dibromide. This product is the precursor to diphenylacetylene, the next synthetic intermediate in the b series. A further purpose of this experiment is to demonstrate the stereospecific addition of bromine to alkenes. The percentage recovery was also calculated and recorded. On completion of this experiment; my lab partners and I were able to successfully synthesize the second intermediate in the b series of the