This problem has to do with the ideal gas law.

Question

What volume is required to contain 100.0 moles O2 at STP?

Answer 1

At STP, one mole of any ideal gas occupies 22.4 L. What that means is that you can just take the moles and scale it by that conversion factor

100 moles O2 * (22.4 L / mol) = 2240.0 L

Answer 2

When dealing with STP (standard temperature and pressure), we can skirt using the ideal gas law if we recall that under these conditions, 1 mole of any ideal gas = 22.4 liters. Thus,...

100.0 mols O2 x 22.41 L / mol = 2241 L (4 sig. figs.)

Answer 3

To find the volume required to contain 100.0 moles of O2 at STP using the ideal gas law:

Ideal Gas Law: PV = nRT

Where:

1. P (Pressure) = 1 atm (standard pressure)
2. V (Volume) = ?
3. n (Number of moles) = 100.0 moles
4. R (Gas constant) = 0.0821 L·atm/K·mol
5. T (Temperature) = 273.15 K (standard temperature)

Substitute the values into the equation and solve for V:

V = (nRT) / P V = (100.0 moles * 0.0821 L·atm/K·mol * 273.15 K) / 1 atm V = 2239.65 L

Thus, the volume required is 2239.65 liters.