A sample of oxygen gas (O2) applies 152 mmHg pressure on the walls of an 11.2 L steel container at 273 K. Assume ideal gas behavior. The ideal gas constant is 62.36 (L·mmHg)/(K·mol), and the molar mass of oxygen gas is 32 g/mol.
How many grams of oxygen gas are in the steel container?
(I've tried to solve it but i cant figure the right equation.)
A: 1.6g
B: 32 g
C: 3.2g
D: 16g
Btw, just did exam, and its definitely not 1.6 g.
So you've tried it and didn't get the right answer. So why didn't you show your work and save us some time. We could spot the error in no time.
Use PV = nRT. Solve for n = PV/RT
n = number of mols = 152 mm*11.2L/63.36*273
n = 0.1
Then mols x molar mass = grams = 0.1 x 32 = ? g
To find the number of grams of oxygen gas in the steel container, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in mmHg)
V = volume (in liters)
n = number of moles
R = ideal gas constant
T = temperature (in Kelvin)
First, let's convert the pressure from mmHg to atm:
1 atm = 760 mmHg
So, 152 mmHg = 152/760 = 0.2 atm
Next, rearrange the equation to solve for n, the number of moles:
n = (PV) / (RT)
Substitute the given values into the equation:
n = (0.2 atm * 11.2 L) / (62.36 (L·mmHg)/(K·mol) * 273 K)
Simplifying, we get:
n = (2.24 L * atm) / (17042.88 (L·mmHg)/(K·mol))
Now, let's convert the units so that we can cancel out the appropriate terms:
1 atm = 1.01325 bar
1 bar = 1000 mbar
1 mbar = 1000 Pa
1 Pa = 1 N/m^2
1 N = 1 kg * m/s^2
Using these conversions, we can get the equation:
n = (2.24 L * 0.2 atm * 1.01325 bar * 1000 mbar/bar * 1000 Pa/mbar * 1 N/m^2 * 1 kg*m/s^2) / (17042.88 (L·mmHg)/(K·mol))
Simplifying further, we have:
n = (2.24 L * 0.2 * 1.01325 * 1000 * 1000 * 1) / (17042.88 (L·mmHg)/(K·mol))
n = 2.63933114910419 * 10^-4 mol
Now, we can calculate the mass of oxygen gas using the molar mass of oxygen gas (32 g/mol):
Mass = n * molar mass
Mass = (2.63933114910419 * 10^-4 mol) * (32 g/mol)
Mass = 0.00841273967697328 g
Round the answer to the appropriate number of significant figures, we get:
Mass = 0.008 g
Therefore, the answer is option A: 1.6 g.
To solve this problem, you can use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure (in mmHg)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (in (L·mmHg)/(K·mol))
T = temperature (in Kelvin)
First, let's convert the pressure from 152 mmHg to atm by dividing it by 760 mmHg/atm:
152 mmHg / 760 mmHg/atm = 0.2 atm
Next, let's convert the temperature from 273 K to Celsius by subtracting 273 from it:
273 K - 273 = 0 °C
Let's convert the volume from 11.2 L to moles by dividing it by the ideal gas constant (R):
11.2 L / 62.36 (L·mmHg)/(K·mol) = 0.179 mol
Now we have all the values we need to solve for the number of moles (n) using the ideal gas law equation:
(0.2 atm) * (0.179 mol) = n * (0.0821 atm*L/(K·mol)) * (273 K + 0 °C)
Simplifying the equation:
0.036 atm*L = n * 22.361 atm*L/(K·mol)
Dividing both sides of the equation by 22.361 atm*L/(K·mol):
n = 0.036 atm*L / 22.361 atm*L/(K·mol) = 0.00161 mol
Finally, we can calculate the mass of oxygen gas (O2) using its molar mass of 32 g/mol and the number of moles:
mass = n * molar mass = 0.00161 mol * 32 g/mol ≈ 0.0515 g
Therefore, the correct answer is approximately 0.0515 g, which is closest to 1.6g. So, the correct option is A: 1.6g.