Question

A producer gas, used as an industrial fuel gas with the following composition by volume: $H_{2}=28%$, $CO=12%$, $CH_{4}=2%$, $CO_{2}=16%$ and $N_{2}=42%$. Find the volume of air required for the complete combustion of $1~m^{3}$ of this gas [Air contains 21% $O_{2}$ by volume].

• A. \(1.143~m^{3}\)
• B. \(3.005~m^{3}\)
• C. \(2.286~m^{3}\)
• D. \(1.523~m^{3}\)

Hint:

Always ignore \(CO_{2}\) and \(N_{2}\) in combustion calculations as they are already fully oxidized or inert. Focus only on the "hungry" molecules like \(H_{2}\), \(CO\), and Hydrocarbons!

Answer 1

The Correct Option is A

Concept: Combustion calculations are based on stoichiometry. We need to determine the amount of oxygen required for each combustible component in the gas mixture.

Step 1: Write the combustion equations for each component.

• For \(H_{2}\): \(H_{2} + 0.5 O_{2} \rightarrow H_{2}O\) (1 vol \(H_{2}\) needs 0.5 vol \(O_{2}\))
• For \(CO\): \(CO + 0.5 O_{2} \rightarrow CO_{2}\) (1 vol \(CO\) needs 0.5 vol \(O_{2}\))
• For \(CH_{4}\): \(CH_{4} + 2 O_{2} \rightarrow CO_{2} + 2H_{2}O\) (1 vol \(CH_{4}\) needs 2 vol \(O_{2}\)) Note: \(CO_{2}\) and \(N_{2}\) are non-combustible.

Step 2: Calculate total oxygen required per \(1~m^{3}\) of gas.
\[ O_{2, required} = (0.28 \times 0.5) + (0.12 \times 0.5) + (0.02 \times 2) \] \[ O_{2, required} = 0.14 + 0.06 + 0.04 = 0.24~m^{3} \]

Step 3: Calculate the required volume of air.
Since air is 21% oxygen: \[ V_{air} = \frac{O_{2, required}}{0.21} = \frac{0.24}{0.21} \approx 1.1428~m^{3} \] Rounding gives \(1.143~m^{3}\).