Chemistry, 11.04.2020 00:53 kaylijocombs

The active component in one type of calcium dietary supplement is calcium carbonate. A 1.2450 g tablet of the supplement is added to 65.00 mL of 0.4984 M HCl(aq) and allowed to react. After completion of the reaction, the excess HCl(aq) requires 37.15 mL of 0.2065 M NaOH(aq) for its titration to the equivalence point. What is the calcium content of the tablet, expressed in milligrams of Ca?

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Answer from: andrewweltz421

The calcium content of the tablet is 486 mg.

The reaction between the HCl and the supplement occurs as follows;

CaCO3(s) + 2HCl(aq) > CaCl2(aq) + CO2(g) + H2O(l)

Total amount of HCl reacted = 65/100 L × 0.4984 M = 0.032 moles

From the reaction;

HCl(aq) + NaOH(aq) ---> NaCl (aq) + H2O(l)

Number of moles of NaOH = 37.15/1000 × 0.2065 M = 0.0077 moles

Since the reaction is 1:1, 0.0077 M of excess HCl is present

Amount of HCl reacted = 0.032 moles - 0.0077 moles = 0.0243 moles

1 mole of calcium carbonate reacts with 2 moles of HCl

x moles of calcium carbonate reacts with 0.0243 moles of HCl

x = 1 mole × 0.0243 moles/2 moles

x = 0.01215 moles

Mass of CaCO3 reacted = 0.01215 moles × 100 g/mol = 1.215 g

If 100 g of CaCO3 contains 40 g of Ca

1.1215 g of CaCO3 contains 1.215 g × 40 g/100 g = 0.486 g or 486 mg

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Answer from: srice6

$\large \boxed{\text{495.5 mg }}$

Explanation:

This is an example of back titration.

You add a known excess of HCl to the CaCO₃. Some of it reacts with the CaCO₃.

You titrate the left-over HCl.

From the difference you can calculate the amount of CaCO₃.

100.09

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂CO₃

m/g: 1.2450

V/mL: 65.00

c/mol·L⁻¹: 0.4984

NaOH + HCl ⟶ NaCl + H₂O

V/mL: 37.15

c/mol·L⁻¹: 0.2065

1. Total moles of HCl

$\text{Moles of HCl}= \text{65.00 mL HCl} \times \dfrac{\text{0.4984 mmol HCl}}{\text{1 mL HCl}} = \text{32.40 mmol HCl}$

2. Excess moles of HCl

$\text{Moles of HCl} = \text{37.15 mL NaOH} \times \dfrac {\text{0.2065 mmol NaOH}}{\text{1 mL NaOH}} \times \dfrac{\text{1 mmol HCl}}{\text{1 mmol NaOH}}\\\\= \text{7.671 mmol HCl}$

3. Moles of HCl that reacted

n = (32.40 - 7.671) mmol HCl = 24.72 mmol HCl

4. Moles of CaCO₃

$\text{Moles of CaCO}_{3} = \text{24.72 mmol HCl} \times \dfrac{\text{1 mmol CaCO}_{3}}{\text{2 mmol HCl}} = \text{12.36 mmol CaCO}_{3}$

5. Mass of Ca

$\text{Mass of Ca} = \text{12.36 mmol CaCO}_{3} \times \dfrac{\text{1 mmol Ca}}{\text{ 1 mmol CaCO}_{3}} \times \dfrac{\text{40.08 mg Ca}}{\text{1 mmol Ca}}\\\\= \textbf{495.5 mg Ca}\\\\\text{The calcium content of the tablet is $\large \boxed{\textbf{495.5 mg }}$}$

Answer from: Quest

they are all solids.

Answer from: Quest

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