Answer
$P_{C_2H_6}=24.19\ kPa$
Work Step by Step
We consider $100 \mathrm{~kg}$ of this mixture. The mole numbers of each component are
$$
\begin{aligned}
N_{\mathrm{CH} 4} & =\frac{m_{\mathrm{CH} 4}}{M_{\mathrm{CH} 4}}=\frac{70 \mathrm{~kg}}{16 \mathrm{~kg} / \mathrm{kmol}}=4.375\ \mathrm{kmol} \\
N_{\mathrm{C} 2 \mathrm{H} 6} & =\frac{m_{\mathrm{C} 2 \mathrm{H} 6}}{M_{\mathrm{C} 2 \mathrm{H} 6}}=\frac{30 \mathrm{~kg}}{30 \mathrm{~kg} / \mathrm{kmol}}=1.0\ \mathrm{kmol}
\end{aligned}
$$ The mole number of the mixture is
$$ N_m=N_{\mathrm{CH} 4}+N_{\mathrm{C} 2 \mathrm{H} 6}=4.375+1.0=5.375\ \mathrm{kmol}
$$ The mole fractions are $$
\begin{gathered}
y_{\mathrm{CH} 4}=\frac{N_{\mathrm{CH} 4}}{N_m}=\frac{4.375 \mathrm{kmol}}{5.375 \mathrm{kmol}}=0.8139 \\
y_{\mathrm{C} 2 \mathrm{H} 6}=\frac{N_{\mathrm{C} 2 \mathrm{H} 6}}{N_m}=\frac{1.0 \mathrm{kmol}}{5.375 \mathrm{kmol}}=0.1861
\end{gathered}
$$ The final pressure of ethane in the final mixture is $$
P_{\mathrm{C} 2 \mathrm{H} 6}=y_{\mathrm{C} 2 \mathrm{H} 6} P_m=(0.1861)(130 \mathrm{kPa})=\mathbf{2 4 . 1 9\ k P a}
$$
