# A 1.24-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of...

## Question:

A 1.24-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 3.20 g of silver chloride. Identify the metal.

## Alkaline Earth Metal:

The alkaline earth metals are the metals belonging to group 2 of the periodic table. Alkaline earth metals have a combining capacity of 2. The metal can be identified using the molar mass of the metal.

{eq}MCl_2 + 2AgNO_3 \to 2AgCl + M(NO_3)_2 {/eq}

The given mass of alkaline earth metal chloride = 1.2 g

The given mass of silver chloride = 3.20 g

The molar mass of silver chloride = 143.32 g/mol

So, the moles of silver chloride in 3.20 g of silver chloride is,

{eq}\begin{align} \rm Moles \ of \ AgCl & = \rm \frac{3.20 \ g}{143.32 \ g/mol} \\ & = \rm 0.0223 \ mol \end{align} {/eq}

The mole-to-mole ratio of alkaline earth metal chloride and silver chloride is 1:2. So, the moles of alkaline earth metal chloride that reacts to form silver chloride is,

{eq}\begin{align} \rm Moles \ of \ MCl_2 & = \rm \frac{1}{2} \times 0.0223 \ mol \\ & = \rm 0.0111 \ mol \end{align} {/eq}

The molar mass of MCl_2 can be calculated as follows,

{eq}\begin{align} \rm 0.0111 \ mol & = \rm \frac{1.24 \ g}{Molar \ mass \ of \ MCl_2} \\ \rm Molar \ mass \ of \ MCl_2 & = \rm \frac{1.24 \ g}{0.0111 \ mol} \\ & = \rm 111.71 \ g \end{align} {/eq}

The Molar mass of alkaline earth metal chloride is the sum of mass of alkaline earth metal and the mass of chlorine. So, the mass of alkaline earth metal is,

Molar mass of MCl{eq}_2 {/eq} = Molar mass of M + 2(Molar mass of Cl)

111.71 g/mol = Molar mass of M + 2(35.45 g/mol)

111.71 g/mol = Molar mass of M + 70.9 g/mol

Molar mass of M = 111.71 g/mol - 70.9 g/mol = 40.81 g/mol

The alkaline earth metal that has a molar mass of 40.81 g/mol is calcium metal.