Chapter 21, Problem 110SCQ

(a)

Interpretation Introduction

Interpretation:

To estimate the mass of lime required to precipitate magneisum from sea water.

Concept introduction:

Magnesium is an alkaline earth metal and is prsent in sea water. The magnesium present in sea water can be extracted by precipitating it with either $\text{CaO}$ or $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$. The pure metal from the salt obtained can be obtained by the process of electrolysis.

$Numberofmoles=\frac{Mass}{Molecularmass}$

$\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{×}\text{Molecular}\text{mass}$

(b)

Interpretation Introduction

Interpretation:

To estimate the mass of product produced at anode and cathode on electrolysis of ${\text{MgCl}}_{\text{2}}$ and also total number of Faradays of electricity used.

Concept introduction:

Magnesium is an alkaline earth metal and is prsent in sea water. The magnesium present in sea water can be extracted by precipitating it with either $\text{CaO}$ or $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$. The pure metal from the salt obtained can be obtained by the process of electrolysis.

The reduction happens at cathode

The oxidation happens at anode.

$Numberofmoles=\frac{Mass}{Molecularmass}$

$\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{×}\text{Molecular}\text{mass}$

The total number of Faradays of electricity used is calculated by the number of electrons involved in the reaction

(c)

Interpretation Introduction

Interpretation:

To estimate how many Joules are required per mole for the given industrial process

Concept introduction:

Magnesium is an alkaline earth metal and is prsent in sea water. The magnesium present in sea water can be extracted by precipitating it with either $\text{CaO}$ or $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$. The pure metal from the salt obtained can be obtained by the process of electrolysis.

$Numberofmoles=\frac{Mass}{Molecularmass}$

$\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{×}\text{Molecular}\text{mass}$

$\text{1kWh=}\text{1}\text{kilowatt}\text{hour}\text{=}\text{3}\text{.6}\text{×}{\text{10}}^{\text{6}}\text{J}$