How many grams of O2 are contained in a 25.0L sample at 5.20 atm and 28.0C?
168 grams
Use PV = nRT and solve for n = number of mols.
Then n = grams/molar mass. You know molar mass and n, solve for grams.
To calculate the number of grams of O2 in a sample, we can use the ideal gas law equation, which is as follows:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
First, we need to convert the given conditions to Kelvin:
Temperature in Kelvin = 28.0 + 273.15 = 301.15 K
Next, we need to rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
n = (5.20 atm) * (25.0 L) / (0.0821 L·atm/mol·K) * (301.15 K)
n ≈ 2.3332 moles
Now, we can calculate the molar mass of O2, which is 32.00 g/mol.
Finally, to find the grams of O2 in the sample, we multiply the number of moles by the molar mass:
Grams = moles * molar mass
Grams = 2.3332 moles * 32.00 g/mol
Grams ≈ 74.66 grams
Therefore, there are approximately 74.66 grams of O2 in the 25.0 L sample at 5.20 atm and 28.0°C.
To find the number of grams of O2 contained in a sample, we can use the ideal gas law equation: PV = nRT, where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles
R = gas constant = 0.0821 L·atm/mol·K
T = temperature in Kelvin (K)
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 28.0°C + 273.15 = 301.15 K
Now we need to calculate the number of moles (n) using the ideal gas law equation:
n = PV / RT
Substituting the given values:
n = (5.20 atm * 25.0 L) / (0.0821 L·atm/mol·K * 301.15 K)
Simplifying:
n = 1.30 mol
The number of moles of O2 is 1.30 mol.
Next, we need to convert moles to grams using the molar mass of O2, which is 32.00 g/mol:
grams = moles * molar mass
grams = 1.30 mol * 32.00 g/mol
Calculating:
grams = 41.6 g
Therefore, there are 41.6 grams of O2 contained in the 25.0L sample at 5.20 atm and 28.0°C.