Titration problem

Question

some magnesium carbonate was added to 145 mal of 1.00mol L^-1 HCL. When the reaction had finished, the solution was acidic. 25.0mL of 0.500 mol L^-1 Na2CO3 solution was required to neutralize the excess acid. What mass of magnesium carbonate was originally used?

MgCO_{3 (s)} +2HCl_(aq) —> MgCl_2(aq) + CO₂ +H2O_(I)

Na₂CO_{3 (s)} + 2HCl _(aq) —> 2NaCL +CO_2(g) +H2O(l)

Answer 1

Original moles HCl present = 145 ml x 1 L / 1000 ml x 1.00 mo/L = 0.145 moles HCl

2HCl + Na₂CO₃ ==> NaCl + CO₂ + H2O

moles Na₂CO₃ used to back titrate excess HCl = 25 ml x 1 L / 1000 ml x 0.500 mol/L = 0.0125 mol Na₂CO₃

mols HCl back titrated (excess) = 0.0125 mol Na₂CO₃ x 2 mol HCl/mol Na₂CO₃ = 0.025 mol HCl in excess

mols HCl used in the original titration = 0.145 mol - 0.025 mol = 0.120 mol HCl used to react with MgCO₃

Molar mass MgCO3 = 84.3 g / mol

MgCO₃(s) + 2HCl ==> MgCl₂ + CO₂ + H2O

mass MgCO₃ present = 0.120 mol HCl x 1 mol MgCO₃ / 2 mol HCl x 84.3 g MgCO₃/mol = 5.06 g MgCO₃