To determine the proportion of calcite in limestone, the student has performed t

Question

To determine the proportion of calcite in limestone, the student has performed the following experiment:

1) the sample is crushed into powder and is weighed by an analytical scale. The sample weight was 651.3 mg.

2) Powder was quantitatively transferred into Erlenmeyer flask and distilled water was added to stir the powder

3) The student added 12.0 ml of hydrochloric acid, c (HCl) = 1.396 mol L-1

4) After the reaction, the student heated the solution in order to drive CO2 out

5) He cooled the solution to room temperature and excess acid was titrated with NaOH. c (NaOH) = 0.1004 mol L-1. Consumption of base was 41.96 ml.

Q: What is the mass fraction of calcite in limestone? To make this determination was true, it is necessary to fulfill a condition regarding the system of limestone. What is this condition?

Explanation / Answer

Mass of limestone (M) = 651.3 mg = 0.6513 g

Molecular weight of Calcite or CaCO3 (MW) = 100.09 g/mol

Volume of HCl added (V1) = 12.0 mL = 0.012 L

Concentration of HCl (C1) = 1.396 mol/L

Concentration of NaOH (C2) = 0.1004 mol/L

Volume of NaOH consumed (V2) = 41.96 mL = 0.04196 L

Moles of HCl added (m1) = V1 X C1 = 0.012 L X 1.396 mol/L = 0.0167 mol

Moles of unreacted HCl (m2) = V2 X C2 = 0.04196 L X 0.1004 mol/L = 0.0042 mol

Moles of Calcite or CaCO3 (m3) = (m1 - m2) / 2 (Divided by 2 as 2 moles of HCl is required to react with CaCO3 to form CaCl2) = (0.0167 mol - 0.0042 mol) /2 = 0.0062 mol

Mass fraction of calcite = (m3 X MW) / M = (0.0062 mol X 100.09 g/mol) / 0.6513 g = 0.953

The condition is that we should assume that the limestone contains only calcite or calcium carbonate and not magnesium carbonate.

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