Answer
$\dot{m}=21.16\text{ lbm/s}$
Work Step by Step
The molar mass of the mixture is determined from $$
M_m=y_{\mathrm{CH} 4} M_{\mathrm{CH} 4}+y_{\mathrm{C} 2 \mathrm{H} 6} M_{\mathrm{C} 2 \mathrm{H} 6}=0.95 \times 16+0.05 \times 30=16.70 \mathrm{lbm} / \mathrm{lbmol}
$$ The specific volume of the mixture is $$
v=\frac{R_u T}{M_m P}=\frac{\left(10.73 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{lbmol} \cdot \mathrm{R}\right)(520 \mathrm{R})}{(16.70 \mathrm{lbm} / \mathrm{lbmol})(100 \mathrm{psia})}=3.341\ \mathrm{ft}^3 / \mathrm{lbm}
$$ The volume flow rate is $$
\dot{V}=A V=\frac{\pi D^2}{4} V=\frac{\pi(36 / 12 \mathrm{ff})^2}{4}(10 \mathrm{ff} / \mathrm{s})=70.69\ \mathrm{ft}^3 / \mathrm{s}
$$ and the mass flow rate is $$
\dot{m}=\frac{\dot{v}}{v}=\frac{70.69 \mathrm{ft}^3 / \mathrm{s}}{3.341 \mathrm{ft}^3 / \mathrm{lbm}}=\mathbf{2 1 . 1 6\ l b m / s}
$$
