Chapter 20, Problem 14PS

Interpretation Introduction

Interpretation:

The mass of $\text{Mg}$ produced from $\text{1}\text{.0}\text{L}$ of sea water and the volume of sea water which is needed to prepare $\text{100}\text{.0}\text{Kg}\text{of}\text{}\text{Mg}$ should be determined.

Concept introduction:

• $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation established in accordance with the Law of conservation of mass.
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• $\begin{array}{l}\text{Mass}\text{=}\text{Density}\text{×Volume}\\ \\ \text{Volume}\text{=}\frac{\text{Mass}}{\text{Density}}\end{array}$
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Electrolysis is a chemical decomposition produced by passing an electric current through a liquid or solution contains ion.

Answer to Problem 14PS

$\text{0}\text{.3354}\text{g}$ of $\text{Mg}$ can be produced from $\text{1}\text{.0}\text{L}$ of sea water

$\text{297}\text{.41}\text{L}$ of sea water will be needed to produce $\text{100}\text{.0}\text{Kg}\text{of}\text{}\text{Mg}$.

Explanation of Solution

Given,

Treatment of sea water with $Ca{\left(OH\right)}_2$ gives insoluble $Mg{\left(OH\right)}_2$. This reacts with $HCl$ to produce $MgC{l}_2$, which is then dried, and the metal $\left(Mg\right)$ is obtained by electrolysis of the molten salt.

Thus,

The sea water which contains magnesium ions and this ion is reacts with $Ca{\left(OH\right)}_2$ gives the insoluble $Mg{\left(OH\right)}_2$ along with $C{a}^{2+}ion$

$Mg{\left(OH\right)}_2$ When reacts with aqueous hydrochloric acid, $MgC{l}_2$ will produced in good yield. After the electrolysis of the molten salt of $MgC{l}_2$, magnesium metal ($Mg$ ) can be isolated.

Balanced net ionic equation for the above reaction can be written as,

  $\begin{array}{l}\text{Ca}{\left(\text{OH}\right)}_{\text{2(s)}}^{}{\text{+Mg}}_{\text{(aq)}}^{\text{2+}}\to \text{Mg}{\left(\text{OH}\right)}_{\text{2(s)}}\text{+}{\text{Ca}}_{\text{(aq)}}^{\text{2+}}\\ \\ \text{Mg}{\left(\text{OH}\right)}_{\text{2(s)}}\text{+}{\text{2H}}_{\text{3}}{\text{O}}_{\text{(aq)}}^{\text{+}}\to {\text{Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+4H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ {\text{MgCl}}_{\text{2(l)}}\to {\text{Mg}}_{\text{(s)}}{\text{+Cl}}_{\text{2(g)}}\end{array}$

We are assuming that there are $\text{1}\text{.025}\text{g}\left(\text{1}\text{.025}\text{g}\text{=}\text{1}\text{.0}\text{L}\text{×}\text{1}\text{.025}{\text{g/cm}}^{\text{3}}\right)\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ present in 1L of sea water.

From the balanced equation above mentioned, the stoichiometric ratio between $Ca{\left(OH\right)}_2$ and $\text{Mg}$ is $1:1$

The mass of magnesium obtained from 1L of sea water can be calculated as follow,

    $\begin{array}{l}\frac{\text{1}\text{.025}\text{g}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{74}\text{.093}\text{g/mol}}\text{=}\text{0}\text{.0138}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\\ \\ \text{Mass}\text{of}\text{Mg}\text{obtained}\text{=}\text{0}\text{.0138}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{×}\frac{\text{1}\text{mol}\text{Mg}}{\text{1}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}\text{×}\frac{\text{24}\text{.305}\text{g}}{\text{1}\text{mol}\text{Mg}}\\ \\ \text{=}\text{0}\text{.3354}\text{g}\end{array}$

$\text{0}\text{.3354}\text{g}$ of $\text{Mg}$ can be produced from $\text{1}\text{.0}\text{L}$ of sea water

$\begin{array}{l}\text{The amount of magnesium in}\text{100}\text{Kg}\text{=}\frac{\text{100000}\text{g}}{\text{24}\text{.305}\text{g/}\text{mol}}\\ \\ \text{=}\text{}\text{}\text{}\text{4114}\text{.37}\text{mol}\end{array}$

The stoichiometric ratio between the reactant and the product Mg is 1:1, therefore the volume of sea water required to produced 100 kg of $\text{Mg}$ can be calculated as follows,

$\begin{array}{c}\text{4114}\text{.37}\text{mol}\text{Mg}\text{×}\frac{\text{1}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{1}\text{mol}\text{Mg}}\text{×}\frac{\text{74}\text{.093}\text{g}}{\text{1}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}\text{=}\text{304846}\text{.016}\text{g}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\\ \\ \text{Volume}\text{of}\text{Sea}\text{water}\text{to}\text{produce}\text{100}\text{Kg}\text{Mg}\text{=}\frac{\text{304846}\text{.016}\text{g}}{\text{1}\text{.025}{\text{g/cm}}^{\text{3}}}\\ \\ \text{=}\text{297410}\text{.74}{\text{cm}}^{\text{3}}\\ \\ \text{=}\text{297}\text{.41}\text{L}\end{array}$

$\text{297}\text{.41}\text{L}$ of sea water will be needed to produce $\text{100}\text{.0}\text{Kg}\text{of}\text{}\text{Mg}$.

Conclusion

The mass of $\text{Mg}$ produced from $\text{1}\text{.0}\text{L}$ of sea water and the volume of sea water which is needed to prepare $\text{100}\text{.0}\text{Kg}\text{of}\text{}\text{Mg}$ was determined.