c
(c) $1.12 \,mL$ is obtained from $4.12\, mg$ $\therefore$ $22400\, mL$ will be obtained from $\frac{{4.12}}{{1.12}} \times 22400\,mg = 84.2\,g$
$\underset{1\,mol.}{\mathop{ROH}}\,+C{{H}_{3}}MgI\to \underset{1\,mol=22400\,cc}{\mathop{C{{H}_{4}}+Mg<_{I}^{OR}}}\,$