a 500ml sample of oxygen gas at 24 degree celsius was prepared by decomposing a 3% aqueous soution of hydrogen peroxide in the presence of manganese catalyst by the reaction: 2h2o2 = 2 h20 + o2...
the oxygen was colletec by displacement of water. the total pressure of gas collected was 755 mmHG. THe vapoer of water at 24 dgree celsius is 23 mmHG.
A. What is the partial pressure of oxygen gas in the mixture?
b. How many moles of oxygen gas are in the mixture?
Please help? I'm not sure where to start!
To solve this problem, you need to use the ideal gas law and the mole concept.
Step 1: Convert the temperature from Celsius to Kelvin.
Add 273 to the Celsius temperature to convert it to Kelvin.
Temperature in Kelvin = 24°C + 273 = 297 K
Step 2: Calculate the partial pressure of water vapor.
Given that the vapor pressure of water at 24°C is 23 mmHg, the partial pressure of water vapor in the mixture is 23 mmHg.
Step 3: Calculate the partial pressure of oxygen gas.
Total pressure - partial pressure of water vapor = partial pressure of oxygen gas
Partial pressure of oxygen gas = 755 mmHg - 23 mmHg = 732 mmHg
Step 4: Convert the partial pressure of oxygen gas to atmospheres.
Partial pressure of oxygen gas in atmospheres = 732 mmHg / 760 mmHg/atm = 0.963 atm
Step 5: Apply the ideal gas law to calculate the number of moles of oxygen gas.
PV = nRT
n = (PV) / (RT)
Where:
P = partial pressure of oxygen gas in atmospheres = 0.963 atm
V = volume of the gas in liters = 500 mL = 0.500 L
R = ideal gas constant = 0.0821 L.atm/mol.K
T = temperature in Kelvin = 297 K
n = (0.963 atm * 0.500 L) / (0.0821 L.atm/mol.K * 297 K)
n = 0.0242 moles
Therefore, the partial pressure of oxygen gas in the mixture is 732 mmHg, and there are approximately 0.0242 moles of oxygen gas in the mixture.
To calculate the partial pressure of oxygen gas in the mixture, you need to subtract the vapor pressure of water from the total pressure of the gas. Here's how to do it:
A. Partial pressure of oxygen gas:
1. Subtract the vapor pressure of water from the total pressure of the gas: Partial pressure of oxygen gas = Total pressure of gas - Vapor pressure of water.
Partial pressure of oxygen gas = 755 mmHg - 23 mmHg.
Partial pressure of oxygen gas = 732 mmHg.
So, the partial pressure of oxygen gas in the mixture is 732 mmHg.
Now let's move on to calculating the number of moles of oxygen gas.
B. Number of moles of oxygen gas:
1. Convert the total pressure of the gas to atm by dividing by 760 mmHg (1 atm = 760 mmHg):
Total pressure of gas in atm = 755 mmHg / 760 mmHg.
Total pressure of gas in atm = 0.9934 atm.
2. Use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
3. Convert the volume of the oxygen gas from milliliters (ml) to liters (L) by dividing by 1000:
Volume of oxygen gas = 500 ml / 1000.
Volume of oxygen gas = 0.5 L.
4. Convert the temperature from degrees Celsius (°C) to Kelvin (K):
Temperature in Kelvin = 24 °C + 273.15.
Temperature in Kelvin = 297.15 K.
5. Rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT.
6. Substitute the values into the equation:
n = (0.9934 atm) * (0.5 L) / (0.0821 L * atm/mol * K) * (297.15 K).
n = 0.0208 mol.
So, there are approximately 0.0208 moles of oxygen gas in the mixture.
PO2 = Ptotal - PH2O vapor.
Use PV = nRT to solve for n = moles O2 gas.
For P use the partial pressure of O2 from the a part. Don't forget to convert T to kelvin.