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)}](https://tex.z-dn.net/?f=2CH_4_%7B%28g%29%7D%20%2B%203O_2_%7B%28g%29%7D%5Crightarrow%202CO_%7B%28g%29%7D%20%2B%204H_2O_%7B%28g%29%7D)
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