Respuesta :
Answer:
Volume of carbon dioxide = 2.47 L
Explanation:
Given that:
For methane gas:-
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25.0 + 273.15) K = 298.15 K
V = 2.80 L
Pressure = 1.65 atm
Using ideal gas equation as:
[tex]PV=nRT[/tex]
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
Applying the equation as:
1.65 atm × 2.80 L = n ×0.0821 L atm/ K mol × 298.15 K
⇒n for methane gas = 0.1887 mol
For oxygen gas:-
Temperature = 31 °C
T = (31 + 273.15) K = 304.15 K
V = 35.0 L
Pressure = 1.25 atm
Using ideal gas equation as:
[tex]PV=nRT[/tex]
Applying the equation as:
1.25 atm × 35.0 L = n ×0.0821 L atm/ K mol × 304.15 K
⇒n for oxygen gas = 1.7521 mol
According to the reaction:-
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
1 mole of methane gas reacts with 2 moles of oxygen gas
Also,
0.1887 mole of methane gas reacts with 2*0.1887 moles of oxygen gas
Moles of oxygen gas = 0.3774 moles
Available moles of oxygen gas = 1.7521 moles
Limiting reagent is the one which is present in small amount. Thus, methane gas is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of methane gas on reaction forms 1 mole of carbon dioxide
Thus,
0.1887 mole of methane gas on reaction forms 0.1887 mole of carbon dioxide
Moles of carbon dioxide = 0.1887 moles
Given that:- Pressure = 2.50 atm
Temperature = 125 °C
T = (125 + 273.15) K = 398.15 K
Using ideal gas equation as:
[tex]PV=nRT[/tex]
Applying the equation as:
2.50 atm × V = 0.1887 moles ×0.0821 L atm/ K mol × 398.15 K
Volume of carbon dioxide = 2.47 L
