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Calculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of CO_(2) can be obtained by heating 1 kg of 90 % pure magnesium carbonate. |
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Answer» Solution :The balanced chemical equation is `MgCO_(3)rarr` MgO + `CO_(2)` MOLAR mass of `MgCO_(3)` is 84 g `mol^(-1)` . 84 g `MgCO_(3)`contain 24 g of Magnesium. `:.`100 g of `MgCO_(3)` contain = ` (24 g MG)/(84 gMgCO_(3)) xx 100 g MgCO_(3) ` = 28.57 g Mg i.e. percentage of magnesium = 28.57 % . 84 g `MgCO_(3)` contain 12 g of carbon  `:. ` 100g `MgCO_(3)` contain `(12 g C)/(84 g MgCO_(3)) xx 100 g MgCO_(3)` = 14.29 g of carbon `:. ` percentage of carbon =14.29% 84 g `MgCO_(3)` contains 48 g of oxygen `:.` 100 g `MgCO_(3) `contains `(48 g O)/(84 g MgCO_(3)) xx 100g MgCO_(3)` = 57.14 g of oxygen `:.` percentage of oxygen = 57.14% As per the stoichiometric equation, 84 g 100% pure `MgCO_(3)` on heating gives 44 g of `CO_(2)` `:. ` 1000 g of 90% pure `MgCO_(3)` gives `(44 g)/(84 g xx 100%)xx 90% xx 1000`g = 471.43 g of `CO_(2)` = 0.471 kg of `CO_(2)` |
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