A 0.6739 g sample of a pure carbonate, X,CO,(s), was dissolved in 50.0 mL of 0.1850 M HCl(aq). The excess HCl(aq)...

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Answer 1

Answer #1

Question 1.

(i) Moles of HCl react with the carbonate = {(50/1000) L * 0.185 mol/L} - {(24.7/1000) L * 0.098 mol/L} = 0.0068294 mol

(ii) The identity of the cation (X): Molar mass of X_nCO₃ = 0.6739 g/(0.0068294 mol) = 98.676 g/mol; Cation (X) = Ca

Note: The molar mass of CaCO₃ = 100 g/mol

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