91Ó°ÊÓ

Calculate the number of moles of HCl:

$\begin{array}{rcl} Moles of HCl& =& 0.1 L so\ln ×\frac{0.75 mol HCl}{L so\ln }\\                                           & =& 0.075 mol.\\ Moles of HCl  used& =& 0.0750 mol-0.0125 mol\\                                           & =& 0.0625 mol.\end{array}$

On multiplying the mole number of HCl used by molar ratio, we get:

$\begin{array}{rcl}Moles of Mg& =& 0.0625 mol HCl×\frac{1 mol Mg}{2 mol HCl}\\                            & =& 0.03125 mol  Mg.\backslash \mathrm{hfill}\\ Mass of Mg& =& 0.03125 mol Mg×\frac{24.31 g Mg}{1 mol Mg}\\                             & =& 076 g.\end{array}$

Divide the mass of Ag to the mass of the impure metal, then multiply by 100:

$\begin{array}{rcl}Mass \% of Mg& =& \frac{0.76 g Mg}{1.32 g impure metal}×100\\                                 & =& 57.6\%.\end{array}$