Question

A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution:

$\mathit{M}\mathit{g}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{H}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{M}\mathit{g}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

After 1.32 g of the impure metal was treated with 0.100 L of 0.750 M HCl, 0.0125 mol of HCl remained. Assuming the impurities do not react, what is the mass % of Mg in the sample?

Short answer

The mass percentage of Mg in the sample is 57.6 % Mg.

Step-by-step solution

Finding the moles of HCl used

Calculate the number of moles of HCl:

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

Finding the mass percentage of Mg

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

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

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

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