a 500ml sample of o2 gas at 24 degrees celsius was prepared by decomposing a 3% aqueous solution of hydrogen peroxide, in this presence of a small amount of manganese catalyst by the reaction 2H2O2>2H2O+O2. The oxygen thus prepared was collected by displacement of water. The total pressure of gas collected was 755mm Hg. What is the partial pressure of O2 in the mixture? How many moles of O2 are in the mixture? (The vapor pressure of water at 24 degrees C is 23mm Hg.)
To find the partial pressure of O2 in the mixture, we need to consider Dalton's Law of Partial Pressures. According to this law, the total pressure of a mixture of gases is the sum of the partial pressures of each gas.
Given:
Total pressure of gas collected = 755 mmHg
Vapor pressure of water at 24 degrees Celsius = 23 mmHg
To find the partial pressure of O2, we need to subtract the vapor pressure of water from the total pressure:
Partial pressure of O2 = Total pressure - Vapor pressure of water
= 755 mmHg - 23 mmHg
= 732 mmHg
Therefore, the partial pressure of O2 in the mixture is 732 mmHg.
To find the number of moles of O2 in the mixture, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = partial pressure of O2
V = volume of the O2 gas (500 mL or 0.5 L)
n = number of moles of O2
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (convert 24 degrees Celsius to Kelvin: 24 + 273 = 297 K)
Rearranging the equation, we get:
n = PV / RT
Substituting the values:
n = (732 mmHg * 0.5 L) / (0.0821 L·atm/(mol·K) * 297 K)
Performing the calculations, we find:
n ≈ 15 moles
Therefore, there are approximately 15 moles of O2 in the mixture.
To find the partial pressure of O2 in the mixture, we need to subtract the vapor pressure of water from the total pressure of the gas collected.
Partial pressure of O2 = Total pressure of gas collected - Vapor pressure of water
Given:
Total pressure of gas collected = 755 mm Hg
Vapor pressure of water at 24 degrees Celsius = 23 mm Hg
Partial pressure of O2 = 755 mm Hg - 23 mm Hg
Partial pressure of O2 = 732 mm Hg
Therefore, the partial pressure of O2 in the mixture is 732 mm Hg.
To find the number of moles of O2 in the mixture, we can use the ideal gas equation:
PV = nRT
Where:
P = Partial pressure of gas (in atm)
V = Volume of gas (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 volume of the O2 gas sample from mL to L.
Volume of O2 gas sample = 500 mL = 500/1000 L = 0.5 L
Now, we can rearrange the ideal gas equation to solve for the number of moles (n):
n = PV / RT
Let's plug in the values:
n = (732 mm Hg) * (0.5 L) / (0.0821 L.atm/mol.K * (24°C + 273.15))
n = (732 mm Hg * 0.5 L) / (0.0821 L.atm/mol.K * 297.15 K)
n = 37.77 / 24.218415
n = 1.558 moles (rounded to 3 decimal places)
Therefore, there are approximately 1.558 moles of O2 in the mixture.
Ptotal = pO2 + pH2O
You know Ptotal and pH2O; solve for pO2.
Find moles from PV = nRT.