Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 20, Problem 12PS
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.

Expert Solution & Answer
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Answer to Problem 12PS

0.3354g of Mg can be produced from 1.0L of sea water

297.41L of sea water will be needed to produce 100.0KgofMg.

Explanation of Solution

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)2×24.305g1molMg =0.3354g

0.3354g of Mg can be produced from 1.0L of sea water

The amount of magnesium in100Kg=100000g24.305g/mol =4114.37mol

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 Mg can be calculated as follows,

4114.37molMg×1molCa(OH)21molMg×74.093g1molCa(OH)2=304846.016gCa(OH)2VolumeofSeawatertoproduce100KgMg =304846.016g1.025g/cm3 =297410.74cm3=297.41L

297.41L of sea water will be needed to produce 100.0KgofMg.

Conclusion

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

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