To find the mass of oxygen withdrawn from the cylinder, we start by calculating the number of moles left in the cylinder after some oxygen is withdrawn. We use the ideal gas law in the form:

$ n = \frac{PV}{RT} $

Substituting the given values:

$ P = 11 $ atm converts to $ 11 \times 1.01 \times 10^5 \, \text{N/m}^2 $

$ V = 30 \, \text{liters} = 30 \times 10^{-3} \, \text{m}^3 $

$ R = \frac{100}{12} \, \text{J/mol K} $

$ T = 27^\circ \text{C} = 300 \, \text{K} $

We calculate the moles after oxygen has been withdrawn:

$ n = \frac{12 \times 1.01 \times 10^5 \, \text{N/m}^2 \times 30 \times 10^{-3} \, \text{m}^3}{\left(\frac{100}{12}\right) \times 300} $

Simplifying the expression:

$ n = \frac{12 \times 1.01 \times 12}{10} = 14.54 \, \text{moles} $

Next, determine the moles of oxygen removed:

$ \text{Moles removed} = 18.20 - 14.54 = 3.656 \, \text{moles} $

Finally, convert the moles removed into mass:

$ \text{Mass removed} = 3.656 \times 32 = 116.99 \, \text{g} = 0.116 \, \text{kg} $

Thus, the mass of the oxygen withdrawn from the cylinder is approximately 0.116 kg.