Part II 1. Antacid Brand 2. Mass of Whole Tablet (g) 3. Mass of Crushed Tablet a
ID: 1029607 • Letter: P
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Part II 1. Antacid Brand 2. Mass of Whole Tablet (g) 3. Mass of Crushed Tablet and Boat (g) . Mass of Boat after Tablet removed (g) 5. Mass of Tablet added to 200mL acid (g) Tums 1.284 1.514 0.231 1.283 If needed Trial 3 Part IlI 6. Stomach Acid (HCI) used (mL) 7. Initial Buret Reading (mL) 8. Final Buret Reading (mL) 9. Volume NaOH added (mL) 10. Average Volume NaOH Used (mL) Trial 1 24.0 3.01 10.89 7.88 Trial 2 24.0 10.89 18.25 7.36 7.62 If needed Trial 3 Part IV: Trial 1 23.0 19.5 20.41 0.91 Trial 2 25.0 20.41 21.85 1.44 11. Volume filtered acid added to flask (mL) 12. Initial Buret Reading (NaOH) (mL) 13. Final Buret Reading (NaOH) (mL) 4. Volume NaOH Added (mL) 15. Volume HCI Remaining in sample (mL) 16. Average HCI Remaining in 25 mL (mL) 17. Volume HCI Neutralized in 200mL Sample (mL) 18. Volume HCI Neutralized by Whole Tablet(mL) 2.746719 4.724409 3.735564304 170.1154856 170.1154856 Post Lab Questions 1. Ifthe HCI used in this experiment was prepared at a concentration of0.150M. Using your ca tablet, calculate the mass of active ingredient (CaCO3) in the tablet in milligrams.Explanation / Answer
The volume of HCl neutralized by the tablet = 170.1154856 mL.
The concentration of the stomach HCl = 0.150 M. Therefore, the mole(s) of HCl added = (170.1154856 mL)*(1 L/1000 mL)*(0.150 M) = 0.025517 mole.
Write down the balanced chemical equation for the reaction between CaCO3 and HCl as
CaCO3 (s) + 2 HCl (aq) -------> CaCl2 (aq) + CO2 (g) + H2O (l)
As per the stoichiometric equation,
1 mole CaCO3 = 2 moles HCl.
Mole(s) of CaCO3 neutralized by 0.025517 mole HCl = (0.025517 mole HCl)*(1 mole CaCO3/2 moles HCl) = 0.0127585 mole.
Molar mass of CaCO3 = (1*40.078 + 1*12.011 + 3*15.999) g/mol = 100.086 g/mol.
Mass of CaCO3 neutralized = (0.0127585 mole)*(100.086 g/mol) = 1.27694 g = (1.27694 g)*(1000 mg/1 g) = 1276.94 mg (ans).
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