Question

25.
Two ores X and Y were taken. On heating these ores it
Y were taken. On heating these ores it was observed that
(a) ore X gives CO2 gas, and
(b) ore Y gives SO, gas.
Write steps to convert these ores into metals, giving chemical equations
of the reactions that take place.​

Answer 1

Answer:

ore x is carbonate ore whereas ore y is sulphide ore

following are the steps to convert these ores into pure metal :_

1. Concentration of ore :-  in this process the unwanted materials are removed from the ores
2. For sulphide ore :-

             Roasting (heating of ore in presence of oxygen) -

             for example ore of mercury

            Cinnabar [HgS]  is the ore of mercury . when it is heated in air ,                                                           firstly it is converted into mercuric oxide[HgO] . Mercuric oxide is reduced    to mercury by heating .

 2HgS(s) + O2(g) = 2HgO(s) + SO2(g)

 2HgO(s) (on heating in presence of O2)  = 2Hg(l) + O2(g)                  

   2. For carbonate ore :-

                   Calcination (heating in absence of oxygen)

                    for example ore of zinc

                    Sphalerite (zinc blende) is the ore of zinc  

                   ZnCO3 (s) (on heating) = ZnO(s) + SO2(g)

                   ZnCO3(s) + C(s) = Zn(s) + C(g)

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Answer 2

Answer:- Two ores are given X and Y.

On heating these ores x gives co2 and y give so2.

It means that x is metal carbonate and y is metal sulphide.

So,by calcination

Metal carbonate heated strongly in the presence of limited supply of air it gives us:- MCo3 ---------> MO + Co2-----(I)

Heat

Now by roasting

Metal sulphide heated in the presence of excess air,then it gives us:-

MSo3 ------->MO + So2----(ii)

From equation 1 & 2

Metal oxide are reduced to metal by most reducing agent called carbon.

MO + c------->m +Co

(Coke)

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