A single aluminum carbonate tablet weighs 2.48g. How many grams of AlCl3 would be produced by the reaction of a single tablet?
Al2(CO3)3 + 6 HCl = 2 AlCl3 + 3 CO2 + 3 H2O the ratio between Al2(CO3)3 and HCl is 1 : 6 moles Al2(CO3)3 required = 2.48 x 1/6=0.401 moles Al2(CO3)3 = 2.48 g /233.987 g/mol=0.0106 the ratio between Al2(CO3)3 and AlCl3 is 1 : 2 moles AlCl3 = 0.0106 x 2=0.0212 mass AlCl3 = 0.0212 mol x 133.34 g/mol=2.83 g
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