Calcium carbonate is heated in a test tube and it decomposes to produce CO2 that is collected over water in the following setup. The volume of the gas that is collected is 0.278 L and the total pressure in the collection tube is 1.87 atm at 21 degrees C. The vapor pressure of the water at 21 degrees C is 18.7 tor. What mass of CO2 was collected in the collection tube?
Ptotal = pH2O + pCO2
Ptotal = 1.87 atm x (760 mm/1 atm) = 1421.2 mm
1421.2 mm = 18.7 mm + pCO2
Solve for pCO2. That is the pressure in mm for the dry gas.
Then PV = nRT
You know PCO2, V = 0.278 L; n = ?; R = you know; T = 273 + 21 = 294
Solve for n = mols CO2
Then mols CO2 x molar mass CO2 = grams CO2.
Post your work if you get stuck or have questions.
Okay this is what I did initially but I was not too sure of it. Thank you very much
To find the mass of CO2 collected, we need to use the Ideal Gas Law equation and consider the partial pressure of CO2 in the collection tube.
The Ideal Gas Law equation is given by:
PV = nRT
Where:
P = total pressure in the collection tube
V = volume of the gas collected
n = number of moles of the gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 21 °C + 273.15
T(K) = 294.15 K
Next, we need to calculate the partial pressure of CO2 by subtracting the vapor pressure of water from the total pressure.
Partial pressure of CO2 = Total pressure - Vapor pressure of water
Partial pressure of CO2 = 1.87 atm - 18.7 torr
To convert torr to atm, we use the conversion factor:
1 torr = 1 mmHg
1 atm = 760 mmHg
1 atm = 760 torr
So, the vapor pressure of water in atm is:
18.7 torr * (1 atm / 760 torr) = 0.0247 atm
Partial pressure of CO2 = 1.87 atm - 0.0247 atm
Partial pressure of CO2 = 1.8453 atm
Now, we can rearrange the Ideal Gas Law equation to solve for moles of CO2:
n = PV / RT
Substituting the values:
n = (1.8453 atm * 0.278 L) / (0.0821 L.atm/mol.K * 294.15 K)
n = 0.015797 mol
Finally, we can calculate the mass of CO2 using the molar mass of CO2.
Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O)
Molar mass of CO2 = 44.01 g/mol
Mass of CO2 = n * molar mass
Mass of CO2 = 0.015797 mol * 44.01 g/mol
Mass of CO2 = 0.6972 g (rounded to four decimal places)
Therefore, the mass of CO2 collected in the collection tube is approximately 0.6972 grams.
To find the mass of CO2 collected in the collection tube, we can use the ideal gas law equation and the concept of partial pressure.
The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the given pressure to atm and the given temperature to Kelvin.
Given:
Volume (V) = 0.278 L
Total pressure (P) = 1.87 atm
Temperature (T) = 21 °C (convert to Kelvin)
To convert Celsius to Kelvin, add 273.15:
T = 21 °C + 273.15 = 294.15 K
Now, we need to consider the partial pressure of water vapor in the collection tube. Since the collected gas is being collected over water, we need to subtract the vapor pressure of the water from the total pressure to obtain the actual pressure of the CO2.
Vapor pressure of water (Pvap) = 18.7 torr
We need to convert torr to atm:
Pvap = 18.7 torr * (1 atm / 760 torr) = 0.0247 atm
Now we can calculate the actual pressure of the CO2:
P_actual = P - Pvap
P_actual = 1.87 atm - 0.0247 atm = 1.8453 atm
Next, let's rearrange the ideal gas law equation to solve for the number of moles of CO2 (n):
n = PV / RT
n = (1.8453 atm * 0.278 L) / (0.0821 L·atm/mol·K * 294.15 K)
n ≈ 0.0234 mol (rounded to four decimal places)
Since 1 mole of CO2 has a molar mass of approximately 44 g/mol, we can find the mass of CO2 using the equation:
Mass = n * Molar mass
Mass = 0.0234 mol * 44 g/mol
Mass ≈ 1.03 g (rounded to two decimal places)
Therefore, the mass of CO2 collected in the collection tube is approximately 1.03 grams.