In the lab, 5.8 L of methane gas is burned with excess oxygen gas at SATP. How much carbon dioxide gas is produced?
**Based on the balanced chemical equation the ratio of methane gas to carbon dioxide is 1:1, this being said, does it make sense that as 5.8 L of methane gas is burned that 5.8 L of carbon dioxide would be produced??
Yes, as long as the pressure/temperature of the CH4 is the same as the CO2 being produced.
To calculate the amount of carbon dioxide gas produced, we need to use stoichiometry, which involves balancing the chemical equation and using the molar ratio.
The balanced chemical equation for the combustion of methane (CH4) with oxygen gas (O2) is:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that one molecule of methane (CH4) reacts with two molecules of oxygen (O2) to produce one molecule of carbon dioxide (CO2) and two molecules of water (H2O).
Now, let's convert the volume of methane to moles using the ideal gas law, which states:
PV = nRT
Where:
P = pressure (SATP = 101.3 kPa)
V = volume (5.8 L)
n = number of moles (unknown)
R = ideal gas constant (8.314 J/(K·mol))
T = temperature (SATP = 273.15 K)
Rearranging the equation, we get:
n = PV / RT
n = (101.3 kPa * 5.8 L) / (8.314 J/(K·mol) * 273.15 K)
Calculating the value of n, we find the number of moles of methane gas.
Next, since methane and carbon dioxide have a 1:1 molar ratio in the balanced equation, the number of moles of carbon dioxide produced will be the same as the number of moles of methane.
Finally, to find the volume of carbon dioxide gas at SATP, we can use the ideal gas law again:
V = nRT / P
Where:
V = volume (unknown)
n = number of moles (same as the number of moles of methane)
R = ideal gas constant (8.314 J/(K·mol))
T = temperature (SATP = 273.15 K)
P = pressure (SATP = 101.3 kPa)
By substituting the values into the equation, we can calculate the volume of carbon dioxide gas.