17.5 mL of oxygen gas were collected at room temperature 22C and 100.2 kPa of atmospheric pressure.
a. How many moles of oxygen gas were produced?
b.What is the molar volume of the oxygen gas at the conditions in the laboratory?
note, how do you change kpa to atm?
sorry about the other post.
To solve these problems, we will use the ideal gas law equation:
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) or 8.314 J/(mol·K))
T = temperature in Kelvin
a. To find the number of moles of oxygen gas produced, we need to convert the given conditions to Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 22 °C + 273.15
T(K) ≈ 295.15 K
Second, let's convert the pressure from kilopascals (kPa) to atmospheres (atm):
1 atm = 101.325 kPa (approximately)
P(atm) = P(kPa)/101.325
P(atm) = 100.2 kPa / 101.325
P(atm) ≈ 0.9896 atm
Now we can plug the values into the ideal gas law equation:
PV = nRT
(0.9896 atm)(17.5 mL) = n(0.0821 L·atm/(mol·K))(295.15 K)
Note: We need to convert the volume from milliliters (mL) to liters (L):
1 L = 1000 mL
(0.9896 atm)(17.5 mL / 1000 mL) = n(0.0821 L·atm/(mol·K))(295.15 K)
0.01729 = n(0.0821 L·atm/(mol·K))(295.15 K)
Simplifying further:
0.01729 = n(24.217015)
Divide both sides by 24.217015 to solve for n:
n = 0.01729 / 24.217015
n ≈ 0.000714 moles
Therefore, approximately 0.000714 moles of oxygen gas were produced.
b. To find the molar volume of the oxygen gas, we can use the ideal gas law and rearrange it to solve for V:
V = (nRT)/P
P = 0.9896 atm (from part a)
n = 0.000714 moles (from part a)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 295.15 K (converted temperature)
V = (0.000714 mol)(0.0821 L·atm/(mol·K))(295.15 K) / 0.9896 atm
Simplifying further:
V ≈ 0.0211 L
Therefore, the molar volume of the oxygen gas at the given conditions is approximately 0.0211 L.
To convert kilopascals (kPa) to atmospheres (atm), divide the pressure in kilopascals by approximately 101.325 (which is a conversion factor between kPa and atm).
To find the number of moles of oxygen gas produced, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles (mol)
R = ideal gas constant, which is 0.0821 L·atm/(mol·K)
T = temperature in Kelvin (K)
Let's begin by converting the given temperature from Celsius to Kelvin:
T (K) = T (°C) + 273.15
T (K) = 22°C + 273.15
T (K) = 295.15 K
Now, let's convert the pressure from kilopascals (kPa) to atmospheres (atm):
1 atm = 101.325 kPa
To convert kPa to atm, divide the given pressure by 101.325:
P (atm) = 100.2 kPa / 101.325
P (atm) ≈ 0.99 atm
Next, we can calculate the number of moles of oxygen gas:
n = PV / RT
n = (0.99 atm) * (17.5 mL / 1000) / (0.0821 L·atm/(mol·K)) * 295.15 K
n ≈ 0.007 moles
Therefore, there are approximately 0.007 moles of oxygen gas produced.
To find the molar volume of the oxygen gas at the conditions in the laboratory, we can use the equation:
Molar volume = Volume / Moles
The given volume is 17.5 mL, but for consistency, we need to convert it to liters:
Volume (L) = Volume (mL) / 1000
Volume (L) = 17.5 mL / 1000
Volume (L) = 0.0175 L
Now, we can calculate the molar volume:
Molar volume = 0.0175 L / 0.007 moles
Molar volume ≈ 2.5 L/mol
Therefore, the molar volume of the oxygen gas at the given conditions is approximately 2.5 L/mol.
To change kPa to atm, divide the pressure in kilopascals by 101.325, as mentioned earlier. In this case, we converted 100.2 kPa to approximately 0.99 atm.
1 atmosphere = 101.325 kPa.
Use PV = nRT
Convert 22 C to Kelvin and use V in Liters. Solve for n = number of moles.
b. The molar volume is the volume occupied by one mole of the gas at STP (pressure of 1 atm and T of 273). But the problem asks for the molar volume of oxygen at the conditions in the laboratory. If you don't have new numbers for the P and T in your lab, I assume the problem means to use 22 C and 100.2 kPa for the room values. Substitute those values and n=1 and solve for V.