What volume of O2 gas, measured at 297K, is needed to completely burn all the methane (CH4) in a 3.00L container at the same T and P?
Write the balanced equation.
CH4 + 2O2 ==> CO2 + 2H2O
Two ways to do this:
a) brute force.
Use PV = nRT to calculate moles CH4 in the 3.00 L container. There is no P listed but you can choose a convenient one (I think 1 atm would be easy). Just remember to use the same P later.
Using the coefficients in the balanced equation convert moles CH4 to moles O2.
Now use PV = nRT again to calculate volume of O2.
b)simple approach. Only works IF both materials are gases, as in this case.
Since T and P are the same, we don't convert everything to moles. Volumes are proportional. So we reason that 3.00 L of CH4 will require (from the coefficients in the equation) twice that or 6.00 L O2. Problem finished. I really like this approach better but students sometimes are confused by it AND they sometimes use this shortcut when the shortcut is not allowed. Good luck.
How do I find n?
The first time or the second time?
Two ways to do this:
a) brute force.
Use PV = nRT to calculate moles CH4 in the 3.00 L container. There is no P listed but you can choose a convenient one (I think 1 atm would be easy). Just remember to use the same P later.
PV = nRT first time
P = 1 atm
V = 3.00 L
R = 0.08206
T = 297 K
Solve for n.
n= 0.12 or close to that. You need to go through it for the actual number. And don't round until all of the problem is done.
Using the coefficients in the balanced equation convert moles CH4 to moles O2.
0.12 moles CH4 x (2 moles O2/1 mole CH4) = 0.12 x 2/1 = 0.24 moles O2. Again, you need to go through to do it exactly.
Now use PV = nRT again to calculate volume of O2.
PV = nRT second time.
P = 1 atm
V = ??
n = 0.24 or whatever you get exactly.
R = you know
T = 297.
Solve for V.
You may have meant how to determine n for the simple way and the answer is "that's what makes it simple." You don't find n; therefore, there is no need to go through PV = nRT twice.
To find the volume of O2 gas needed to completely burn all the methane (CH4) in a 3.00L container at the same temperature and pressure, we need to use the balanced chemical equation for the reaction between methane and oxygen.
The balanced chemical equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O
From this equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
First, we need to calculate the number of moles of methane in the 3.00L container. To do this, we use the ideal gas law equation:
PV = nRT
Where:
P = pressure (constant)
V = volume
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature
Assuming constant pressure and using the ideal gas law equation, we can rearrange it to solve for the number of moles:
n = PV / RT
Since the volume (V), pressure (P), and temperature (T) are given, we can substitute these values into the equation to calculate the number of moles of methane.
Next, we use the stoichiometry of the balanced chemical equation to determine the number of moles of oxygen required. For every 1 mole of methane, we need 2 moles of oxygen.
Finally, we convert the number of moles of oxygen into the volume of oxygen gas using the ideal gas law equation:
V = nRT / P
Substituting the values of the number of moles of oxygen (from the previous calculation), the ideal gas constant, the temperature, and the pressure, we can calculate the volume of oxygen gas required to completely burn all the methane in the 3.00L container.