Answer :
Answer:
[tex]\large \boxed{\text{495.5 mg }}[/tex]
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
[tex]\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}[/tex]
2. Excess moles of HCl
[tex]\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}[/tex]
3. Moles of HCl that reacted
n = (32.40 - 7.671) mmol HCl = 24.72 mmol HCl
4. Moles of CaCO₃
[tex]\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}[/tex]
5. Mass of Ca
[tex]\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 $\large \boxed{\textbf{495.5 mg }}$}[/tex]
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
Learn more: https://brainly.com/question/9743981