Question

An anhydrous sample of CaCO3 (100.089 g/mol) weighing 0.9849 g was quantitatively transferred into a 250.0 mL volumetric flask half-full of Nanopure H2O. One milliliter of 12M HCl was added & the solution was swirled. Then the flask was filled to volume with Nanopure H2O. What is the molarity (M) of the solution?

Answer 1

The molarity of CaCO₃ in the final solution is equal to 0.015 M.

Explanation:

When we add HCl over CaCO₃ we have the following chemical reaction:

CaCO₃ + 2 HCl → CaCl₂ + H₂CO₃

H₂CO₃ (is not stable) → CO₂ + H₂O

number of moles = mass / molecular weight

number of moles of CaCO₃ = 0.9849 / 100.089 = 0.0098 moles

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of HCl = 12 × 0.001 = 0.012 moles

From the reaction we see that 1 mole of CaCO₃ will react with 2 moles of HCl so the limiting reactant will be HCl. Knowing this we formulate the following reasoning:

if         1 mole of CaCO₃ will react with 2 moles of HCl

then   X moles of CaCO₃ will react with 0.012 moles of HCl

X = (0.012 × 1) / 2 = 0.006 moles of CaCO₃

The number of moles of CaCO₃ which remain unreacted are equal to 0.0098 (initial quantity) - 0.006 (reacted with HCl) = 0.0038 moles.

Now the molarity of CaCO₃ in the final solution:

molar concentration = number of moles / volume (L)

molar concentration of CaCO₃ = 0.0038 / 0.250 = 0.015 M

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