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If 1.25 grams of pure CaCO3 required 25.5 mL of a HCl solution for complete reation what is the molarity for the HCl solution?
Wiki User
∙ 14y ago
When calcium carbonate (CaCO3) reacts with hydrochloric acid (HCl) the reaction is:- CaCO3 + 2 HCl -> CaCl2 + H2O + CO2 The weight for 1 mole of calcium carbonate is the atomic weight in grams or 40.8 + 12.011 + 3 x 15.9994 = 100.8092 grams 1.25 grams of CaCO3 is therefore 1.25/100.8092 or .0124 mole There must have been twice this many moles of HCl in the solution or 0.0248 mole. The molarity of the solution is the number of moles per litre so with 0.0248 in 25.5 ml there must be .97 mole in 1 litre. The HCl solution must therefore be .97 molar
Wiki User
∙ 14y ago
Wiki User
∙ 16y agoI'm not exactly sure but I think the answer is 3.425 grams of CaCO3.
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