## Chemistry 12th Edition

1.01 mol $Cl_{2}$
$Si(s)+2Cl_{2}(g)$ -> $SiCl_{4}(l)$ mol of $Cl_{@}$= $\frac{(0.507molSiCl_{4})\times(2molCl_{2})}{1molSiCl_{4}}$ = 1.01 mol $Cl_{2}$