The number of grams of oxygen in $32.2 \mathrm{g}$ of ${\mathrm{Na}}_2{\mathrm{SO}}_4·10{\mathrm{H}}_2\mathrm{O}$ is approximately

${\mathrm{C}}_2{\mathrm{H}}_4$ can react with ${\mathrm{H}}_2$ in presence of a catalyst to form ${\mathrm{C}}_2{\mathrm{H}}_6$ as per the following reaction, ${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\left( \mathrm{g}\right)\stackrel{ \mathrm{Catalyst} }{⟶}{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)$ The amount of ${\mathrm{C}}_2{\mathrm{H}}_4$ in grams required to produce $50$ grams of ${\mathrm{C}}_2{\mathrm{H}}_6$ is