Given the reaction below

$SiC{l}_4(l)+2Mg(s)Si(s)+2MgC{l}_2(s)$

And considering we have $225g$ of $SiC{l}_4$ and $225g$ of $Mg$

Can someone explain what the $2$ below represents? I know SiCl $4$ is the limiting reactant. Trying to apply this method

on another exercise.

How much excess reactant remains?

Mass of Mg consumed:

$m_{\text{c}\text{o}\text{n}\text{s}\text{u}\text{m}\text{e}\text{d}}(Mg)=n_{\text{c}\text{o}\text{n}\text{s}\text{u}\text{m}\text{e}\text{d}}(Mg)M(Mg)=2n(Si)M(Mg)$

$m_{\text{c}\text{o}\text{n}\text{s}\text{u}\text{m}\text{e}\text{d}}(Mg)1\mathrm{.}32$mol$24\mathrm{.}31g=64\mathrm{.}2g$

Mass of $Mg$ remaining:

$m_{\text{r}\text{e}\text{m}\text{a}\text{i}\text{n}\text{i}\text{n}\text{g}}(Mg)=m_{\text{i}\text{n}\text{i}\text{t}\text{i}\text{a}\text{l}}(Mg)-m_{\text{c}\text{o}\text{n}\text{s}\text{u}\text{m}\text{e}\text{d}}(Mg)=225g-64\mathrm{.}2g=160\mathrm{.}8g$