Question

Methane, the principal component of natural gas, is used for heating and cooking. The combustion process is

CH₂(g) + 20₂(g) → CO₂(g) + 2 H₂O (1)
If 19.7 moles of CH4 react with oxygen, what is the volume of CO₂ produced at 24.3 °C and 0.957 atm? Round your answer to 3 significant digits.
Note: Reference the Fundamental constants table for additional information.

Answer 1

V = 502 L

Step-by-step explanation:

Gas Stoichiometry

We're given:

• 
  `n_(CH_4)=19.7\hspace{4}moles`
• 
  `T = 24.3^(\circ)C`
• 
  `P=0.957\hspace{4}atm`
• 
  `V_(CO_2)=?`

CH₄(g) + 2O₂(g) → CO₂(g) + 2 H₂O(l)

First, determine the moles of CO₂ using this ratio:

`(n_(CH_4))/(1)=(n_(CO_2))/(1)\\\\n_(CO_2)=n_(CH_4)*1\\\\n_(CO_2)=19.7\hspace{4}moles`

Now, to determine its volume, use the ideal gas law:

`PV=nRT`

⇒ Isolate V

`V=(nRT)/(P)`

⇒ Convert given temperature to Kelvin

`T = 297.45\hspace{4}K`

`V=((19.7)(0.08206)(297.45))/(0.957)\\\\V=502.45801\hspace{4}L`

⇒ Round to significant figures

`V=502\hspace{4}L`

Therefore, the volume of CO₂ produced is 502 L.