Answer:
Maximum volume of carbon monoxide that can be produced = <u>2.206 × 10² L </u>
Explanation:
<u>Given</u>: Volume of CH₄ = 6.62 × 10² L, and Volume of O₂ = 3.31 × 10² L
To calculate the number of moles of methane and oxygen, we use the ideal gas equation: PV= nRT or n = PV ÷ (RT)
Here, the standard temperature (T) and pressure (P) is 273.15 K and 1 atm, respectively and the ideal gas constant (R) = 0.08206 L·atm.mol⁻¹ K⁻¹
Therefore, the number of moles of CH₄ = (1 atm × 6.62 × 10² L) ÷ (0.08206 L·atm.mol⁻¹ K⁻¹ × 273.15 K) = 29.55 moles
and the number of moles of O₂ = (1 atm × 3.31 × 10² L) ÷ (0.082 L·atm.mol⁻¹ K⁻¹ × 273.15 K) = 14.77 moles
<u>Given reaction</u>: 2CH₄ (g) + 3O₂ (g) → 2CO (g) + 4H₂O (g)
In this partial combustion, 2 moles of methane reacts with <u>3 moles of oxygen to give 2 moles of carbon monoxide.</u>
Since <u>oxygen is the limiting reagent</u>.
Therefore, methane reacts with <u>14.77 moles of oxygen</u> to give (14.77 × 2 ÷ 3) moles of carbon monoxide = <u>9.85 moles of carbon dioxide.</u>
Therefore, <u>volume of carbon dioxide produced</u>= nRT ÷ P = (9.85 mole× 0.082 L·atm.mol⁻¹ K⁻¹× 273.15 K) ÷ 1 atm = 220.62 L = <u>2.206 × 10² L</u>
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<u>Therefore, the</u><u> maximum volume of carbon monoxide</u><u> that can be produced is </u><u>2.206 × 10² L</u><u>.</u>