Question

Limestone consists mainly of the mineral calcite, CaCO3. The carbonate content of 0.5426 g of powdered limestone was measured by suspending the powder in water, adding 10.00 mL of 1.360 M HCl, and heating to dissolve the solid and expel CO2. CaCO3(s) + 2 H + → Ca2+ + CO2↑ + H2O Calcium carbonate FM 100.087 The excess acid required 39.59 mL of 0.1039 M NaOH for complete titration to a phenolphthalein end point. Find the weight percent of calcite in the limestone. ____ wt%

Answer 1

HCl + NaOH    ---------------> > NaCl + H2O

Moles of excess HCl = moles of NaOH

moles of NaOH = 39.59 x 0.1039 / 1000

                          = 4.113 x 10^-3 mol

Initial moles of HCl = 1.360 x 10 / 1000

                              = 0.0136 mol

Moles of reacted HCl = 0.0136 - 4.113 x 10^-3

                                  = 9.487 x 10^-3
CaCO3 + 2 HCl -------------------> CaCl2 + CO2 + H2O

Moles of CaCO3 = 1/2 x moles of reacted HCl

                           = 1/2 x 9.487 x 10^-3

                            = 4.743 x 10^-3 mol

Mass of CaCO3 = = 4.743 x 10^-3 x 100.087

                            = 0.474 g

Weight% = (0.474 /0.5426 ) x 100%

Weight%   = 87.42 %