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Electrolysis of molten MgCl₂ is the final production step in the isolation of magnesium from seawater by the Dow process (Section 22.4). Assuming that 45.6 g of Mg metal forms,(c) How many amps will produce this amount in 3.50 h?

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shaikhmohd0119

Current produced =28.731 A

Dow's Process

 Overall Reaction :

  Mg⁺² + 2Cl⁻ → Mg(l) + Cl₂ (g)

Mass of Mg = 45.6 g

number of moles of Mg = [tex]\frac{45.6}{24.305}[/tex]

                                       = 1.876 moles of Mg

Now ,

       1 mole Mg⁺² = 2 Faraday

       1.876 mole Mg⁺²  = 2× 1.876 Faraday

                                    = 3.752 Faraday

                                    = 3.752 × 96485 C

                                    = 362011.72 C

Now ,

  Given time = 3.50 h = 3.50 × 3600 sec

                     = 12600 sec

Now , we know that

   Charge(Q) = Current(I) × Time (T)

Hence,

        I = [tex]\frac{Q}{T}[/tex]

           =[tex]\frac{362011.72 C}{12600 sec}[/tex]

          = 28.731 A

    Thus from the above conclusion we can say that amount of current produced is 28.731 A

Learn More about Electrolysis here : brainly.com/question/14964810

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