The following equation represents the partial combustion of methane, ch4. 2ch4(g) + 3o2(g) → 2co(g) + 4h2o(g) at constant temperature and pressure, what is the maximum volume of carbon monoxide that can be obtained from 6.62 × 102 l of methane and 3.31 × 102 l of oxygen?

Maximum volume of carbon monoxide that can be produced = 2.206 × 10² L        

Explanation:

Given: 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

Given reaction: 2CH₄ (g) + 3O₂ (g) → 2CO (g) + 4H₂O (g)

In this partial combustion, 2 moles of methane reacts with 3 moles of oxygen to give 2 moles of carbon monoxide.

Since oxygen is the limiting reagent.

Therefore, methane reacts with 14.77 moles of oxygen to give (14.77 × 2 ÷ 3) moles of carbon monoxide = 9.85 moles of carbon dioxide.

Therefore, volume of carbon dioxide produced= nRT ÷ P = (9.85 mole× 0.082 L·atm.mol⁻¹ K⁻¹× 273.15 K) ÷ 1 atm = 220.62 L = 2.206 × 10² L

Therefore, the maximum volume of carbon monoxide that can be produced is 2.206 × 10² L.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

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