Question

A 0.2027 gram sample of finely powdered limestone (mainly CaCO₃) was dissolved in 50
mL of 0.1035 M HCl. The solution was heated to expel CO₂ produced by the reaction. The
remaining HCl was then titrated with 0.1018 M of NaOH and it required 16.62 mL. Calculate
the percentage of CaCO₃ in the limestone sample?

Answer 1

Answer:

1 answer: Neporo4naja [7] 7 months ago 8 0 Answer: 85.9% Explanation: There are two reactions to take into consideration for this problem: The first one is the reaction of limestone with HCl: 1) CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O While the second one is the titration of HCL with NaOH: 2) HCl + NaOH → H₂O + NaCl So first we calculate the number of HCl moles added: 50 mL ⇒ 0.050 L 0.050 L * 0.1035 M = 5.175x10⁻³ moles HCl A number of those 5.175x10⁻³ moles reacted with limestone while the remaining excess reacted with NaOH. The number of NaOH moles used in the titration is: 0.01662 L * 0.1018 M = 1.692x10⁻³ mol NaOH We know from equation 2) that the moles of NaOH are equal to the moles of HCl in the titration process. That means that the number of HCl moles that reacted with CaCO₃ in the limestone sample is: 5.175x10⁻³ moles HCl - 1.692x10⁻³ mol HCl = 3.483x10⁻³ mol HCl Now we use that number of HCl in equation 1) to calculate the number of CaCO₃ moles, which we then convert to grams using its molecular weight: 3.483x10⁻³ mol HCl * * 100 g/mol = 0.1741 g CaCO₃. Finally we calculate the CaCO₃ percentage in the sample: 0.1741 g / 0.2027 g * 100% = 85.9%

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