   Chapter 20, Problem 14PS

Chapter
Section
Textbook Problem

Calculate the mass of Mg that could be obtained by the process described in Study Question 13 from 1.0 L of seawater. To prepare 100. kg of Mg, what volume of seawater would be needed? (The density of seawater is 1.025 g/cm3.)

Interpretation Introduction

Interpretation:

The mass of Mg produced from 1.0L of sea water and the volume of sea water which is needed to prepare 100.0KgofMg should be determined.

Concept introduction:

• Numberofmole=GivenmassofthesubstanceMolarmass
• 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.
• Mass=Density×VolumeVolume=MassDensity
• 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.
Explanation

Given,

Treatment of sea water with Ca(OH)2 gives insoluble Mg(OH)2. This reacts with HCl to produce MgCl2, which is then dried, and the metal (Mg) is obtained by electrolysis of the molten salt.

Thus,

The sea water which contains magnesium ions and this ion is reacts with Ca(OH)2 gives the insoluble Mg(OH)2 along with Ca2+ion

Mg(OH)2 When reacts with aqueous hydrochloric acid, MgCl2 will produced in good yield. After the electrolysis of the molten salt of MgCl2, magnesium metal (Mg ) can be isolated.

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

Ca(OH)2(s)+Mg(aq)2+Mg(OH)2(s)+Ca(aq)2+Mg(OH)2(s)+2H3O(aq)+Mg(aq)2++4H2O(l)MgCl2(l)Mg(s)+Cl2(g)

We are assuming that there are 1.025g(1.025g=1.0L×1.025g/cm3)ofCa(OH)2 present in 1L of sea water.

From the balanced equation above mentioned, the stoichiometric ratio between Ca(OH)2 and Mg is 1:1

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

1.025gCa(OH)274.093g/mol =0.0138molCa(OH)2MassofMgobtained=0.0138molCa(OH)2×1molMg1molCa(OH)2×24

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