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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