Given PV=nRT and the following equation:
CH4+2O2=CO2+2H2O
Calculate the maximum number of ML of CO2 at 750 torr and 295.15 degrees K that can be formed from above conbution of methane ( CH4 ) if 450 mL of methane at 705 torr and 283.15 degrees K are mixed with 725 mL of O2 at 680 torr at 353.15 degrees K
Use PV = nRT and convert CH4 to moles.
Do the same for O2.
This is a limiting reagent problem; determine the limiting reagent.
Use the limiting reagent as a single stoichiometry problem to calculate the number of mols CO2 produced, then use PV = nRT to convert to L. Multiply by 1000 to convert to mL.
To calculate the maximum number of milliliters (mL) of CO2 that can be formed from the given reaction, we need to use the ideal gas law equation PV = nRT and apply stoichiometry.
First, let's calculate the number of moles of methane (CH4) and oxygen (O2) using the ideal gas law. We can then determine which reactant is the limiting reagent.
Methane (CH4) calculations:
P = 705 torr = 705/760 atm (convert to atmosphere)
V = 450 mL = 450/1000 L (convert to liters)
T = 283.15 K
Using PV = nRT, we can rearrange the equation to solve for n:
n = PV / RT
nCH4 = (705/760 atm) * (450/1000 L) / (0.0821 L*atm/(mol*K)) * (283.15 K)
Now, let's calculate the number of moles of oxygen (O2):
P = 680 torr = 680/760 atm (convert to atmosphere)
V = 725 mL = 725/1000 L (convert to liters)
T = 353.15 K
nO2 = (680/760 atm) * (725/1000 L) / (0.0821 L*atm/(mol*K)) * (353.15 K)
Next, we determine the stoichiometry of the reaction by comparing the coefficients of the balanced equation:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane (CH4) produces 1 mole of carbon dioxide (CO2). Therefore, the number of moles of carbon dioxide formed is equal to the number of moles of methane.
Assuming the reaction goes to completion, the number of moles of carbon dioxide formed is nCO2 = nCH4.
Let's calculate nCO2 by comparing the moles of CH4 and O2:
nCH4 = (705/760 atm) * (450/1000 L) / (0.0821 L*atm/(mol*K)) * (283.15 K)
nO2 = (680/760 atm) * (725/1000 L) / (0.0821 L*atm/(mol*K)) * (353.15 K)
Since nCH4 is less than nO2, CH4 is the limiting reagent.
To find the maximum number of moles of CO2, we use nCO2 = nCH4.
Finally, we use the ideal gas law to calculate the volume (V) of CO2 produced:
P = 750 torr = 750/760 atm (convert to atmosphere)
T = 295.15 K
nCO2 = nCH4
VCO2 = nCO2 * RT / P
VCO2 = nCH4 * RT / P
VCO2 = (nCH4) * (0.0821 L*atm/(mol*K)) * (295.15 K) / (750/760 atm)
The maximum number of mL of CO2 that can be formed is VCO2 multiplied by 1000 to convert to milliliters:
Maximum number of mL of CO2 = VCO2 * 1000 mL
Solve for VCO2 by plugging in the previously calculated values to get your final answer.