Answer
$8.28\text{ kg}$
$21.72\text{ kg}$
Work Step by Step
From the equilibrium diagram (Fig. 16-21) at $T=84 \mathrm{~K}$, the oxygen mole fraction in the vapor phase is $34 \%$ and that in the liquid phase is $70 \%$. That is,$$
y_{f, \mathrm{O} 2}=0.70 \text { and } y_{g, \mathrm{O} 2}=0.34
$$ The mole numbers are $$
\begin{aligned}
N_{\mathrm{O} 2} & =\frac{m_{\mathrm{O} 2}}{M_{\mathrm{O} 2}}=\frac{30 \mathrm{~kg}}{32 \mathrm{~kg} / \mathrm{kmol}}=0.9375\ \mathrm{kmol} \\
N_{\mathrm{N} 2} & =\frac{m_{\mathrm{N} 2}}{M_{\mathrm{N} 2}}=\frac{40 \mathrm{~kg}}{28 \mathrm{~kg} / \mathrm{kmol}}=1.429\ \mathrm{kmol} \\
N_{\text {total }} & =0.9375+1.429=2.366\ \mathrm{kmol}
\end{aligned}
$$ The total number of moles in this system is $$
N_f+N_g=2.366
$$ The total number of moles of oxygen in this system is $$
0.7 N_f+0.34 N_g=0.9375
$$ Solving equations (1) and (2) simultneously, we obtain $$
\begin{aligned}
& N_f=0.3696 \\
& N_g=1.996
\end{aligned}
$$ Then, the mass of oxygen in the liquid and vapor phases is
$$
\begin{aligned}
& m_{f, \mathrm{O} 2}=y_{f, \mathrm{O} 2} N_f M_{\mathrm{O} 2}=(0.7)(0.3696 \mathrm{kmol})(32 \mathrm{~kg} / \mathrm{kmol})=\mathbf{~8 . 2 8} \mathbf{k g} \\
& m_{g, \mathrm{O} 2}=y_{g, \mathrm{O} 2} N_g M_{\mathrm{O} 2}=(0.34)(1.996 \mathrm{kmol})(32 \mathrm{~kg} / \mathrm{kmol})=\mathbf{2 1 . 7 2} \mathbf{~k g}
\end{aligned}
$$
