33.0 L of methane (CH4) undergoes complete
combustion at 0.961 atm and 20◦C. How much
CO2 is formed?
Answer in units of L
How much H2O is formed?
I shall be happy to critique you thoughts.
I thought that at fist you would put the information into PV=nRT so .961V=(1)(.08206)(20+273). the volume would then be 25.04539583 L. I am not sure though if I am even on the right track.
If you assume that the problem is asking for volume (and not grams) and that the volume of CO2 is at the same P and T, you can work it just like the H2 and O2 problem previously. That saves a lot of time.
OR, you can use PV = nRT, solve for n, and convert n to whatever T and P you want BUT there are no DIFFERENT conditions listed so this seems like a lot of extra work to me. You should get th same answer either way.
so since there is a 1:1 ratio between Ch4 and co2, would the volume of CO2 formed be equal to the volume of methane?
yes.
CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g)
If you do it the other way though, you should get the same answer (with more work).
n = PV/RT = 0.961*33.0/0.08206*293 = about 1.32 and plug that back into
V = nRT/P = 1.32*0.08206*294/0.961 = 33.0255 but it will be exactly 33.0 if I had not rounded the 1.31898 mols to 1.32.
Thank you
To determine the amount of CO2 and H2O formed during the combustion of 33.0 L of methane (CH4), we need to use the balanced chemical equation for the combustion reaction of methane.
The balanced equation for the combustion of methane is as follows:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane (CH4) reacts to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O). Therefore, we need to convert the given volume of methane (33.0 L) to moles and then use the mole ratios from the balanced equation to find the number of moles of CO2 and H2O formed.
Step 1: Convert the volume of methane to moles.
To convert the volume of methane to moles, we need to use the ideal gas law equation:
PV = nRT, where:
P is the pressure (in atm),
V is the volume (in liters),
n is the number of moles,
R is the ideal gas constant (0.0821 L atm/mol K), and
T is the temperature (in Kelvin).
Given:
Volume of methane (V) = 33.0 L
Pressure (P) = 0.961 atm
Temperature (T) = 20°C = 20 + 273 = 293 K
Using the ideal gas law, we can calculate the number of moles of methane:
n = PV / RT
= (0.961 atm) * (33.0 L) / (0.0821 L atm/mol K) * (293 K)
Step 2: Calculate the moles of CO2 and H2O formed.
From the balanced equation, we can see that 1 mole of methane (CH4) reacts to produce 1 mole of CO2 and 2 moles of H2O.
Therefore, the number of moles of CO2 formed will be the same as the number of moles of methane, and the number of moles of H2O formed will be twice that of the number of moles of methane.
Number of moles of CO2 = Number of moles of methane
Number of moles of H2O = (2 * Number of moles of methane)
Step 3: Convert the moles of CO2 and H2O to volume.
To convert moles to volume, we use the ideal gas law equation:
V = nRT / P, where:
V is the volume (in liters),
n is the number of moles,
R is the ideal gas constant (0.0821 L atm/mol K), and
P is the pressure (in atm).
Using the ideal gas law, we can calculate the volume of CO2 and H2O:
Volume of CO2 = (Number of moles of CO2) * (0.0821 L atm/mol K) * (293 K) / (0.961 atm)
Volume of H2O = (Number of moles of H2O) * (0.0821 L atm/mol K) * (293 K) / (0.961 atm)
Now, you can calculate the volume of CO2 and H2O formed by substituting the values obtained in Step 2 into the volume formulas from Step 3.