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

The reaction shows that each mole of carbonate requires 2 moles of HCl for the reaction .

Number of moles of HCl = 0.010 x 1.360 = 0.0136 moles

The reaction of HCl and NaOH requires equimolar concentration of both the solutions.

So the number of moles of NaOH = 0.03959 x 0.1039 = 0.00411 moles

The number of moles of HCl reacted with carbonate = 0.0136 - 0.00411 = 0.00949 moles

So the number of moles of carbonate = 0.00949 / 2 = 0.004745 moles

The amount of calcite = 0.004745 x 100.087 = 0.4749 g

The weight percent of calcite in the limestone = (0.4749 / 0.5426) x 100

= 87.52 %