F. 4 Chemistry Mole Concept－ppi93gk88d的部落格

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F. 4 Chemistry Mole Concept

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請指教指教~~!!1. Which is the cheaper source of sodium carbonate? A. Anhydrous sodium carbonate at $2 per kg B. Hydrated sodium carbonate (Na2CO3．H2O) at $1 per kg2. 20 g of a crystalline salt contain 9.73 g of water of crystallization. If the percentage composition by mass of the salt is 8.11 % of... 顯示更多請指教指教~~!! 1. Which is the cheaper source of sodium carbonate? A. Anhydrous sodium carbonate at $2 per kg B. Hydrated sodium carbonate (Na2CO3．H2O) at $1 per kg 2. 20 g of a crystalline salt contain 9.73 g of water of crystallization. If the percentage composition by mass of the salt is 8.11 % of aluminium, 14.41% of sulphur, 72.07% of oxygen and 5.41% of hydrogen, derive the formula of the salt. (Given the RAM of Al = 27.0 , S = 32.1 , O = 16 , H = 1 , Na = 23.0 , C = 12.0)

最佳解答:

1. For anhydrous sodium carbonate: Cost of pure Na2CO3 = $2/kg For Na2CO3?H2O: Molar mass of Na2CO3 = 23x2 + 12x1 + 16x3 = 106 g/mol Molar mass of Na2CO3?H2O = 106 + 1x2 + 16 = 124 g/mol Cost of pure Na2CO3 = $1 / (106/124) = $1.17/kg Hence, B is the cheaper source. 2. Mole ratio Al : S : O : H = 8.11/27 : 14.41/32.1 : 72.07/16 : 5.41/1 = 0.300 : 0.449 : 4.50 : 5.41 = 1 : 1.5 : 15 : 18 = 2 : 3 : 30 : 36 The empirical formula of Al2S3O30H36. Formula mass of the salt = 27x2 + 32.1x3 + 16x30 + 1x36 = 666.3 Molar mass of H2O = 1x2 + 16 = 18 Let n be the number of moles of water in each mole of the salt. Mole ratio of Al2S3O30H36 : H2O = 20/666.3 : 9.73/18 = 1 : n n(20/666.3) = 1(9.73/18) n = 18 In the hydrated salt: No. of H atoms in water of crystallization = 18 x 2 = 36 No. of O atoms in water of crystallization = 18 No. of H atoms in the part of anhydrous salt = 36 - 36 = 0 No. of O atoms in the part of anhydrous salt = 30 - 18 = 12 The formula of the crystalline salt = Al2S3O12?18H2O = Al2(SO4)3?18H2O