An ironmaking blast furnace produces hot metal of the composition: $\pu{93.6\%}$ $\ce{Fe},$ $\pu{2.1\%}$ $\ce{Si},$ $\pu{3.6\%}$ $\ce{C},$ $\pu{0.7\%}$ $\ce{Mn}.$ The composition of the iron ore is $\pu{78\%}$ $\ce{Fe2O3},$ $\pu{9\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{Al2O3},$ $\pu{1\%}$ $\ce{MnO},$ $\pu{7\%}$ moisture. Coke rate is $\pu{850 kg/thm}.$ Composition of the coke is $\pu{80\%}$ $\ce{C},$ $\pu{10\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{CaO}.$ Enough limestone is charged to make $\pu{45\%}$ $\ce{CaO}$ in slag and composition of limestone is $\pu{100\%}$ $\ce{CaCO3}.$

Calculate (i) Weight of ore used per tonne of hot metal production; (ii) weight of slag produced per tonne of H.M.; (iii) weight of limestone used to produce given slag composition.


In such problems all I need to balance is iron, silica, $\ce{MnO}$ etc. However, in case of $\ce{MnO}$ the solution in my book reads $\ce{MnO}$ from ore equals $\ce{Mn}$ in hot metal plus $\ce{MnO}$ in slag.

My question is for balancing $\ce{MnO}$: why do I need to drag in $\ce{Mn}$ content? The $\ce{Mn}$ in iron ore should be equal to $\ce{Mn}$ in slag. In fact, I feel that any content apart from $\ce{MnO}$ should not be dragged into picture. The rest of the contents like alumina, iron etc. quite comply to this argument. I would require a vivid justification to this.

  • $\begingroup$ I would think for (i) you only need to consider iron, and for (ii) you only need to consider calcium. Does all the iron end up in the hot metal? Does the calcium from coke and from limestone all end up in the slag? $\endgroup$
    – Karsten
    Commented Aug 29, 2019 at 12:04
  • $\begingroup$ For (ii), I would need help determining the coke: ore mass ratio from the coke rate, though. $\endgroup$
    – Karsten
    Commented Aug 29, 2019 at 12:06
  • $\begingroup$ @KarstenTheis Thankyou for your help.It is the Manganese that I am having doubt with.While they write MnO input=MnO output,MnO output equals "Mn in hot metal+MnO in slag."That is what the solution reads.Please explain why is it not MnO in hot metal. $\endgroup$
    – user586228
    Commented Aug 29, 2019 at 14:35

1 Answer 1


"The Mn in iron ore should be equal to Mn in slag." vs "the solution in my book reads MnO from ore equals Mn in hot metal plus MnO in slag."

It seems that the calculation desired involves breaking down the MnO in ore to Mn + O, and putting some of the Mn into the metal; the rest of the Mn goes into slag, as MnO. The hot metal would contain only metallic Mn, not MnO. It shouldn't be a huge adjustment in the answers.

  • $\begingroup$ Please tell me something..while balacing MnO why am I not taking the $O_2$ that needs to be liberated into consideration.If MnO in iron ore give Mn in Hot Metal+MnO in slag.The oxygen should go somewhere.But my solution neglects that completely. $\endgroup$
    – user586228
    Commented Aug 31, 2019 at 7:52
  • $\begingroup$ Iron ore has Fe2O3. The O goes off with C as CO/CO2. The MnO in iron ore has to lose its O just like the Fe2O3 does. Consider Al2O3 and SiO2 to keep their oxygen; likewise H2O just evaporates. The coke has CaO and SiO2: they keep their oxygen. The overall process takes oxygen-containing metallic compounds and coke and limestone turns them into hot metal and slag and CO/CO2 + H2O. Consider the O in MnO (some of it) to go off as CO/CO2, just like the O in Fe2O3. $\endgroup$ Commented Aug 31, 2019 at 13:29

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