liquid methane is fed to a space heater at a rate of 10kg/hr Liquid methanol (CH

Question

liquid methane is fed to a space heater at a rate of 10kg/hr Liquid methanol (CH_3OH) is fed to a space heater at a rate of 10 kg/hr, and is burned with excess air. The product gas is analyzed, and the following dry-basis (Orsat) mole percentages are determined: CH_3OH: 0.84% C0_2:7.1% CO: 2.4% Calculate the overall conversion of methanol Calculate the percent excess air in the process. The following equations will be helpful CH_3OH + 3/2 O_2 implies CO_2 + 2 H_2O CH_3OH + O_2 implies CO + 2 H_2O

Explanation / Answer

a. given mass of methanol = 10 Kg = 10000g

    molar mass of ethanol = 32.04g/mol

    number of moles of ethanol = 10000/32.04 = 312.11 mol

    percentage of mole of methanol converted = 100-0.84 = 99.16%

    the overall conversion of methanol = 99.16 x 312.11 / 100 = 309.49 moles

                                                            = 309.49 x 32.04 = 9916 g = 9.9 kg

b. percent excess air = 100 - 0.84 - 7.1 - 2.4 = 89.66%

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