What volume of a oxygen at STP was collected? oxygen gas is collected over water at a temperaturee of 10 degrees C and a pressure 1.2 atm. The volume of gas plus water vapor collected is 293ml.
To find the volume of oxygen at STP (Standard Temperature and Pressure), you will need to make a few calculations. The key is to apply the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, we have the following information:
- Temperature (T) = 10 degrees Celsius = 10 + 273.15 = 283.15 K
- Pressure (P) = 1.2 atm
- Volume of gas plus water vapor collected (V) = 293 mL
Now, let's calculate the moles of water vapor present:
First, we convert the volume of water vapor to liters:
293 mL = 293 mL ÷ 1000 mL/L = 0.293 L
Next, we use the ideal gas law to find the moles of water vapor:
PV = nRT
(1.2 atm)(0.293 L) = n(0.0821 L•atm/mol•K)(283.15 K)
Simplifying:
0.3516 = n(0.0821)(283.15)
0.3516 = 1.8450915n
Now, solve for n:
n = 0.3516 / 1.8450915
n ≈ 0.1905 mol
Since water vapor is diatomic (H2O), the number of moles of oxygen gas (O2) will be half of the number of moles of water vapor. Therefore, we have:
n(O2) = 0.1905 / 2
n(O2) ≈ 0.09525 mol
Now, we can find the volume of oxygen gas at STP using the same ideal gas law:
PV = nRT
At STP, the temperature is 273.15 K, and the pressure is 1 atm. We want to find the volume (V). We have the number of moles (n) as approximately 0.09525 mol. And the ideal gas constant (R) is 0.0821 L•atm/mol•K.
1 atm * V = 0.09525 mol * 0.0821 L•atm/mol•K * 273.15 K
Solving for V:
V = (0.09525 mol * 0.0821 L•atm/mol•K * 273.15 K) / 1 atm
V ≈ 2.12 L
Therefore, the volume of oxygen at STP collected is approximated to be 2.12 liters.
To determine the volume of oxygen gas collected at STP (Standard Temperature and Pressure), we need to apply the principles of Dalton's Law of Partial Pressures and use the Ideal Gas Law.
Step 1: Convert the temperature from Celsius to Kelvin.
Given: Temperature (T) = 10 degrees Celsius
To convert to Kelvin (K): T(K) = T(C) + 273.15
T(K) = 10 + 273.15
T(K) = 283.15 K
Step 2: Apply Dalton's Law of Partial Pressures to find the pressure of oxygen gas.
Given: Total Pressure (Ptotal) = 1.2 atm
Pressure of water vapor (Pwater) = collected pressure - total pressure
Pwater = 1.2 atm - 1 atm (since the pressure of water vapor at STP is 1 atm)
Pwater = 0.2 atm
Step 3: Calculate the pressure of the oxygen gas.
Pressure of oxygen gas = Total pressure - Pressure of water vapor
Poxygen = Ptotal - Pwater
Poxygen = 1.2 atm - 0.2 atm
Poxygen = 1 atm
Step 4: Apply the Ideal Gas Law to calculate the volume of oxygen gas at STP.
Given: Volume of gas plus water vapor (V) = 293 ml
Pressure of oxygen gas (Poxygen) = 1 atm
Temperature (T) = 283.15 K
Rearrange the Ideal Gas Law equation to solve for volume:
V1/T1 = V2/T2
V2 = (V1 * T2) / T1
V2 = (293 ml * 273.15 K) / 283.15 K
V2 = 284.96 ml (rounded to two decimal places)
So, the volume of oxygen gas collected at STP is approximately 284.96 ml.