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Malachite on calcination gives `rarr ‘A’ +CO_(2) +H_(2)O`. Compound `‘A’` on reduction with carbon gives `rarr CO+ ‘B’`. Here ‘A’ and ‘B’are:A. `Fe_(2),O_(3),Fe`B. `CuO,Cu`C. `CuCO_(3),CuO`D. `MgO,Mg` |
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Answer» Correct Answer - B `Cu(OH)_(2).CuCO_(3)rarr CUOunderset((a))+CO_(2) +H_(2)O` `CUOunderset((b))+CrarrCO+Cu` |
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