SO2 (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C. The partial pressure of CO2 in the container was __________ atm.
4.02
The correct verified answer is 4.02
To determine the partial pressure of CO2 in the container, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First, we need to find the number of moles of CO2. To do this, we can use the given mass and the molar mass of CO2.
The molar mass of CO2 is:
C = 12.01 g/mol
O = 16.00 g/mol (There are two Oxygen atoms in CO2)
Molar mass of CO2 = 12.01 g/mol + (16.00 g/mol x 2) = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
Now we can calculate the number of moles of CO2:
moles of CO2 = mass of CO2 / molar mass of CO2 = 5.00 g / 44.01 g/mol
Similarly, we can calculate the number of moles of SO2, given its mass and molar mass.
Next, we need to convert the volume of the container from milliliters to liters:
volume = 750.0 mL * (1 L / 1000 mL)
Now we can plug the values into the ideal gas law equation to find the total pressure of the container. Since we are interested in the partial pressure of CO2, we can rearrange the equation to solve for P:
P = nRT / V
Substituting the values, we find that the partial pressure of CO2 in the container is equal to the calculated total pressure.
Finally, we can express the answer in atmospheres by using the appropriate conversion factor (1 atm = 101325 Pa), if necessary.
Use PV = nRT and solve for P.
n = grams/molar mass = 5.00/molar mass CO2
Ignore the SO2.