Question

A single charcoal briquette has a mass of 0.1 kg, and is made up of almost entirely carbon. I want to burn that briquette in air in a sealed oven. When the combustion is complete, the gases in the oven will include CO2,O2,and N2, and the mole fraction of CO2 in this exhaust is 5%. How much air is needed (before the combustion)? Give your answer as kg, kg mole, and as a volume in m^3. For the last part, you'll use the ideal gas law. Assume that p= 1 atm, and T=400C

Answer 1

Solution.

The combustion reaction is

C+O₂=CO₂

The amount of oxygen consumed is the same as the amount of carbon dioxide formed.

0.1 kg of charcoal is 100 g, or 100/12 = 8.33 moles. It means that the amount of CO₂ formed is 8.33 moles as well. It represents 5% of total number of moles.

So, the total number of exhaust products is 8.33/0.05 = 166.6 moles. As the amount of oxygen consumed is the same as the amount of carbon dioxide formed, the number of moles of air was the same, namely 166.6 moles.

The molar composition of air is approx. 78% N₂ and 22% O₂.

The number of moles of N₂ is 0.78*166.6 =130 moles; it weights 130*28 = 3640 g; it occupies the volume V = nRT/p = 130*8.31*(400+273)/101325 = 7.2 m³

The number of moles of O₂ is 166.6-130 = 36.6 moles; it weights 36.6*32 = 1171 g; it occupies the volume V = nRT/p = 36.6*8.31*(400+273)/101325 = 2.0 m³

Therefore, the amount of air is 166.6 moles; the mass of air is 3.64+1.17 = 4.81 kg; the volume of air at 400 ⁰C is 2.0+7.2 = 9.2 m³.