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Unformatted text preview: H2O + CO2-2molH/1molH2O 1.58 mol H-1molC/1molCO2 .702 mol C-(C.702H1.58)/.702 = CH2.25 = C4H9 Fuel Analysis: Amount of CO2 produced is of concern. Actual ratio of interest = massCO2/mass fuel-Octane eqn: C8H18 + 25/2O2 9H2O + 8CO2-2C8H18 + 25O2 18H2O + 16CO2-CO2/fuel = 16mol/2mol = 3.1gCO2/1g fuel-other possibilities:-propane: C3H8 + 5O2 = 4H2O + 3CO2-ratio = 3mol/1mol = 3gCO2/1gfuel-NO improvement-methanol: CH4O + 3/2O2 2H2O + CO2-2CH4O + 3O2 4H2O + 2CO2-ratio = 1mol/1mol = 1.3gCO2/1gfuel-GOOD improvement!!! (b/c oxygenated) Percent Yield:- (mass obtained/theoretical mass possible (from LR calculation)) x 100- good yield: greater than/equal to 80% ex: Calculate percent yield if LR calc predicts 234g Al2O3. Actually obtained 187g.-187/234 = .799 = 79.9%- mass OR mols...
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