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Electrolytic reduction of iron ore - Stoichiometry

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the maximum mass of aluminum that can be made from 408 tonnes of alumina, assuming that aluminum is produced by electrolysis.

    2. Relevant equations

    At the cathode: [tex]Al^{3+} (l) + 3e^{-} \Rightarrow Al (l)[/tex].

    At the anode: [tex]O^{2-} (l) \Rightarrow O_2 (g) + 4e^{-}[/tex]. Is the oxide ion from aluminum oxide in liquid or gaseous state?

    3. The attempt at a solution

    Is there any limiting reagent of sorts? I have never done any stoichiometric calculations on metallurgy.

    Postscript. How do we input chemistry using LaTeX? (Especially the chemical equations and symbols, e.g. how to make Al not italicized.)
     
    Last edited: Oct 16, 2009
  2. jcsd
  3. Oct 16, 2009 #2

    cepheid

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    Regarding formatting, LaTeX posts on PF are automatically in math mode, which means that all letters are italicized (which is the conventional way to typeset mathematical variables). However, the following LaTeX output (without the spaces in the tex tags) will generate the result given below it:

    [ tex ] \textrm{Al}^{3+}_{(\textrm{l})} +3e^{-} \longrightarrow \textrm{Al}_{(\textrm{l})} [ /tex ]

    [tex] \textrm{Al}^{3+}_{(\textrm{l})} +3e^{-} \longrightarrow \textrm{Al}_{(\textrm{l})} [/tex]

    where the \textrm{} enviroment takes us out of math mode and into normal text mode using the roman font.
     
  4. Oct 16, 2009 #3

    cepheid

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    Regarding the problem, it's been a while since I've done chemistry, but isn't it merely a matter of:

    1. Figuring out how many moles of aluminum oxide correspond to 408 tonnes of it.
    2. Figuring out how many moles of Al are produced per mole of aluminum oxide
    3. Converting that many moles of Al into a mass, assuming all of it is converted?

    Edit, and here's a non LaTeX alternative which might or might not be easier for chemistry (your choice). Use the sup and sub tags for superscripts and subscripts respectively

    This input, without spaces in the tags, produces the output below it:

    Al[ sub ](l)[ /sub ][ sup ]3+[ /sup ] + 3e[ sup ]-[ /sup ] → Al [ sub ] (l) [ /sub ] ​

    Al(l)3+ + 3e- → Al(l)
     
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