Question

Zinc and magnesium metal each react with hydrochloric acid according to the following equations:

$$\begin{gathered} \mathit{Z}{\mathit{n}}_{\left(s\right)}\mathbf{+}\mathbf{2}\mathit{H}\mathit{C}{\mathit{l}}_{\left(aq\right)}\mathbf{\to }\mathit{Z}\mathit{n}\mathit{C}{{\mathit{l}}_{\mathbf{2}}}_{\left(aq\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}\left(g\right)} \\ \mathit{M}{\mathit{g}}_{\left(s\right)}\mathbf{+}\mathbf{2}\mathit{H}\mathit{C}{\mathit{l}}_{\left(aq\right)}\mathbf{\to }\mathit{M}\mathit{g}\mathit{C}{{\mathit{l}}_{\mathbf{2}}}_{\left(aq\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}\left(g\right)} \end{gathered}$$

A 10.00g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156ml of 3.00M silver nitrate to produce the maximum possible amount of silver chloride.

1. Determine the percent magnesium by mass in the original mixture.
2. If 78.0ml of HCl was added, What was the concentration of the ?

Short answer

The percent magnesium by mass in the original mixture is $31.35\%$and the concentration of the $HCl$is$6M$.

Step-by-step solution

Determine the magnesium percent

It is given that the mixture requires 156ml of$3.00M$silver nitrate to form a precipitate. Thus,

$$\begin{gathered} MolesofSilvernitrate=Molarity\times Volume \\ =3.00M\times 0.156L \\ MolesofSilvernitrate=0.468moles \end{gathered}$$

Determine the magnesium percent

The moles of chloride ions are $0.466moles$

The stoichiometric amount of $HClis0.468moles$

The stoichiometric amount of Zinc and moles of Magnesium, the total is$0.468moles$

Mass magnesium percent

$$\begin{gathered} MolesofMagnesium=\frac{Mass}{Atomicmass} \\ =\frac{xg}{24.30g/mol} \\ MolesofZinc=\frac{\left(10-x\right)g}{65.39g/mol} \\ 2\left[\left(\frac{x}{24.30}\right)+\left(\frac{10-x}{65.39}\right)\right]=0.468mol \\ Themasspercentageofmagnesium=\frac{Massofmagnesium}{Massofmixture}\times 100 \\ =\frac{3.135g}{10.00g}\times 100 \\ MasspercentageofMagnesium=31.35\% \end{gathered}$$

Concentration of HCl

$$\begin{gathered} MolesofHClis0.468mol \\ ThevolumeofHClis78.8ml=0.078L \\ MolarityofHCl=\frac{moles}{volume} \\ =\frac{0.468mol}{0.078L} \\ MolartiyofHCl=6M \end{gathered}$$

Conclusion

Therefore, the percent magnesium by mass in the original mixture $31.35\%$is and the concentration of the$HClis6M$.