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Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of CO_(2) can be obtained from 100 Kg of is 90% pure magnesium carbonate |
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Answer» Solution :MOLAR mass of MgCO3 = 84.32 g `mol^(-1)` Percentage of Mg = `24/84.32 xx 100 = 28.46%` Percentage of C =`12/84.32 xx 100` = 14.23% Percentage of `O_(3) =48/84.32 xx 100 = 57.0%` `MgCO_(3) rarr MgO + CO_(2)` 84.32 g of 100% pure `MgCO_(3)` gives 44g of `CO_(2)` `:.100xx10^(3)` g of 100 % pure `MgCO_(3)` gives =`44/84.32 xx 100 xx 10^(3)` = `52.182 xx 10^(3)` g`CO_(2)` 100% pure` MgCO_(3)` gives `52.182 xx 10^(3)gCO_(2)` `:.` 90% pure `MgCO_(3)` will give`(52.182xx10^(3))/100 xx 90` = 46963.8 g `CO_(2)` = 46.96 KG `CO_(2)`. |
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