Initially in the $\mathrm{O}_2$ cylinder,
$$
\begin{aligned}
&V_1=30 \text { litre }=30 \times 10^{-3} \mathrm{~m}^3 \\
&P_1=15 \mathrm{~atm}=15 \times 1.01 \times 10^5 \mathrm{~Pa}, \\
&T_1=27+273=300^{\circ} \mathrm{K}
\end{aligned}
$$
If $n_1$ be the moles of $\mathrm{O}_2$ gas in the cylinder,
$$
\begin{aligned}
& P_1 V_1=n_1 R T_1 \\
\therefore & n_1=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300}=18.253
\end{aligned}
$$
$M=$ molecular weight of $\mathrm{O}_2=32 \mathrm{~g}$; Initial cylinder mass $=m_1=n_1 M=18.253 \times 32=584.1 \mathrm{~g}$
Let, $n_2$ - moles of $\mathrm{O}_2$ left in the cylinder
$$
\begin{aligned}
&\therefore \quad n_2=\frac{P_2 V_2}{R T_2}\left[\text { Where, } V_2=30 \times 10^{-3} \mathrm{~m}^3,\right. \\
&\left.P_2=11 \times 1.01 \times 10^5 \mathrm{~Pa}, T_2=17+273=290^{\circ} \mathrm{K}\right] \\
&=\frac{\left(11 \times 1.01 \times 10^5\right)\left(30 \times 10^{-3}\right)}{8.3 \times 290}=13.847 \\
&\therefore m_2=\text { final mass of } \mathrm{O}_2 \text { in cylinder } \\
&=13.847 \times 32=453.1 \mathrm{~g}
\end{aligned}
$$
$\therefore \quad$ Mass of Oxygen taken out of the cylinder.
$$
m_1-m_2=584.1-453.1=131 \mathrm{~g}
$$