Stoichiometric Air-Fuel Ratio for Combustion of Carbon Monoxide in Air

Stoichiometric air-fuel ratio is the ratio of air to fuel in a combustion process that produces complete combustion with no excess fuel or air. For combustion of carbon monoxide (CO) in air, the stoichiometric air-fuel ratio by volume is 2.38.

Explanation:

Combustion of carbon monoxide (CO) in air can be represented by the following balanced equation:

2CO + O2 → 2CO2

From the balanced equation, it can be seen that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.

To determine the stoichiometric air-fuel ratio, we need to calculate the amount of air required to completely combust 1 mole of CO.

1 mole of CO requires 0.5 mole of O2 for complete combustion (from the balanced equation above). Therefore, the air required for combustion of 1 mole of CO is:

Air required = 0.5 mole O2 + N2 (from air)

Assuming air contains 79% nitrogen (N2) and 21% oxygen (O2) by volume, we can calculate the volume of air required as:

Air required = (0.5/0.21) + (0.79/0.21) x 0.5
Air required = 2.38 volume of air

Therefore, the stoichiometric air-fuel ratio by volume for combustion of carbon monoxide in air is 2.38.

Conclusion:

The correct answer to the given question is option B (2.38). The stoichiometric air-fuel ratio is an important parameter in combustion processes as it affects the efficiency of the process and the emissions produced.