A steel cylinder contains 98.0 moles of oxygen gas at temperature of 38 degree Celsius and a pressure of 740 torr. After some of the oxygen gas has been used, the pressure is reduced to 332 torr at a temperature of 24 degree Celsius. What mass of oxygen gas is removed from the cylinder?

To find the mass of oxygen gas removed from the cylinder, we need to use the ideal gas law equation, which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

First, let's convert the temperatures from Celsius to Kelvin. To do this, we add 273.15 to each given temperature:

Initial temperature (T1) = 38 + 273.15 = 311.15 K
Final temperature (T2) = 24 + 273.15 = 297.15 K

Next, let's convert the pressures from torr to atmospheres (atm). 1 atm = 760 torr:

Initial pressure (P1) = 740 torr / 760 torr/atm = 0.974 atm
Final pressure (P2) = 332 torr / 760 torr/atm = 0.437 atm

Now we can use the ideal gas law equation to find the initial and final number of moles of oxygen gas:

PV = nRT

Initial:
(0.974 atm) * V = (98.0 moles) * (0.0821 L * atm / K * mol) * (311.15 K)

Final:
(0.437 atm) * V = n * (0.0821 L * atm / K * mol) * (297.15 K)

Notice that the volume (V) of the cylinder does not change, as it is not mentioned in the problem. Thus, we can set the two equations equal to each other:

(0.974 atm) * V = (0.437 atm) * V
V cancels out on both sides of the equation.

Then we solve for n:

(0.974 atm) * 98.0 moles = (0.437 atm) * n

Now we can find the final number of moles (n) of oxygen gas:

n = (0.974 atm * 98.0 moles) / (0.437 atm)

n = 217.2 moles

Finally, to find the mass of oxygen gas removed from the cylinder, we use the formula:

mass = n * molar mass

The molar mass of oxygen gas (O2) is approximately 32.00 g/mol.

mass = 217.2 moles * 32.00 g/mol

mass = 6942.4 g

Therefore, the mass of oxygen gas removed from the cylinder is 6942.4 grams.