Consider a 100 mol mixture that is 69.0% methane (CH4) and 31.0% ethane (C2H6).

Question

Consider a 100 mol mixture that is 69.0% methane (CH4) and 31.0% ethane (C2H6). To this mixture is added 29.0% excess air. Of the methane present, 92.10% reacts, 92.20% of which forms carbon dioxide (CO2), and the balance forms carbon monoxide (CO). Of the ethane present, 87.4% reacts, 92.20% of which forms carbon dioxide, and the balance forms carbon monoxide What is the theoretical amount of oxygen required for the fuel mixture? Number mol O What amount of air is added to the fuel mixture? Number mol air How many moles of methane are present in the product gas? Number mol CH

Explanation / Answer

Oxygen require for the mixture of fluel = (127.098+ 94.829) mol

Methane present in the product gas = 5.451 mol

Ethane present in the product gas = 3.902 mol

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