Question

Limestone consists mainly of the mineral calcite, CaCO3 (molar mass = 100.087 g/mol), The carbonate content of 1.4056 g of powdered limestone was measured by suspending the powder in water, adding
15.00 mL of 1.777 M HCl, and heating to dissolve the solid and expel CO2 :
CaCO3 (s) + 2 H+ (aq) -> Ca2+ (aq) + H2O(l) + CO2 (g)
The excess acid required 45.20 mL of 0.1005 M NaOH for comple titration to a phenolphthalein end point. Find the weight percent of calcite in the limestone.

Answer 1

Volume of 1.777 M HCl added = 15.0 mL

Milimles of HCl added = 15 * 1.777

= 26.655

Volume of 0.1005 M NaOH used to titrate HCl = 45.20 mL

Milimoles of NaOH used = 0.1005 * 45.20

= 4.5426

Milimoles of HCl titrated by NaOH = Milimole sof NaOH = 4.5426

Milimoles of HCl consumed by CaCO3 = 26.655 - 4.5426

= 22.1124

Now,

CaCO3 (s) + 2 H+ (aq) -> Ca2+ (aq) + H2O(l) + CO2 (g)

From the reaction, 2 moles H+ reacts with 1 mole CaCO3.

So milimoles of CaCO3 which reacts with 22.1124 milimoles of HCl = 22.1124 /2

= 11.0562 mmoles

= 0.011 moles of CaCO3

Molar mass of CaCO3 = 100.087 g/mol

Mass of CaCO3 = 0.011 * 100.087

= 1.1009 g

Mass of limestone = 1.4056 g

Weight percent of calcite in the limestone = 1.1009 / 1.4056 * 100

= 78.32 %