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How many grams of 80% pure marble stone on calcination can give 14 grams of quick lime? |
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Answer» Solution :On calculation marble stone decomposes to GIVE QUICK lime (CaO) `CaCO_(3) to CaO+CO_(2)` 1 mole of CaO =1 mole of `CaCO_(3)` 56 grams of CaO =100 grams of `CaCO_(3)` 14 grams of CaO=? The WEIGHT calcium carbonate required `=(14)/(56) xx 100` =25 grams Percentage purity is only 80% 80 grams of calcium carbonate =100 grams of marble stone 25 grams of calcium carbonate=? Weight of marble stone to be calcinated `=(25)/(80) xx 100` =31.25 grams |
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