Question

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?

Answer 1

Answer:

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.

Answer 2

Answer:

220.67 L

Explanation:

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

From the equation, at constant P and T, V is directly proportional to moles

Thus, According to the reaction:

$2CH_4_{(g)} + 3O_2_{(g)}\rightarrow 2CO_{(g)} + 4H_2O_{(g)}$

Methane gas and oxygen gas react in 2 : 3 ratio

So, Volume of methane gas = 662 L

Volume of oxygen gas = 331 L

Since, Volume of oxygen gas is less. It is the limiting reagent.

So,

3 L of oxygen gas forms 2 L of CO.

331 L of oxygen gas forms (2/3)*331 L of CO.

Volume of CO obtained = 220.67 L