Part II: Using that infomation I need to answer two questions! PLEASE HELP 1. Th

Question

Part II:

Using that infomation I need to answer two questions! PLEASE HELP

1. The HCl used in this experiment was prepared at a concentration of 0.15M. Using the Volume of HCl Neutralized by your whole table Calculate the mass of active ingredient (CaCO3) in the tablet in milligrams.

Part II:

Antacid Brand Tums Mass of Whole Tablet (g) 1.545 Mass of Crushed Tablet and Boat (g) 1.191 Mass of Boat after Tablet removed (g) 0.22 Mass of Tablet added to 200mL acid (g) 0.971 Part III: Trial 1 Trial 2 Stomach Acid (HCl) used (mL) 25.0 25.0 Initial Buret Reading (mL) 0 0 Final Buret Reading (mL) 24.98 25.28 Volume NaOH added (mL) 24.98 25.28 Average Volume NaOH Used (mL) 25.13 Part IV: If needed Trial 1 Trial 2 Trial 3 Volume filtrated added to flask (mL) 25.0 25.0 25.0 Initial Buret Reading (NaOH) (mL) 0.51 0.73 Final Buret Reading (NaOH) (mL) 19 18.83 Volume NaOH Added (mL) 18.49 18.1 Volume HCl Remaining in sample (mL) 18.39 18.01 Average HCl Remaining in 25 mL (mL) 18.20 Volume HCl Neutralized in 200mL Sample (mL) 54.40 Volume HCl Neutralized by Whole Tablet(mL) 86.55

Explanation / Answer

In a 200ml of sample, we have 0.15M HCl added with 0.971g of antacid

out of 200ml, only 54.40 ml was only neutralized by 0.971g of antacid.

No of moles of HCl in 54.4 ml = n = M*V = 0.15*0.0544 =0.00816 moles = 8.16millimoles

Each mole of CaCO3 can neutralize two moles of HCl

CaCO3 + 2HCl --> CaCl2 + H2CO3

Hence number of moles of CaCO3 required to neutralize 8.16 millimoles of HCl is

ncaco3 = 8.16/2 = 4.08 millimoles

Hence mass of CaCO3 in 0.971gm of antacid = 0.00408*100 = 0.408 gms

Hence fraction of CaCO3 in antacid = 0.408/0.971 = 0.42 or 42%

Hence mass of CaCO3 in whole tablet = 1.545*0.42 =0.649 gms

alternatively, whole tablet can neutralize 86.55 ml of HCl

Hence total moles of HCl = 0.15*0.08655 = 0.0129825

Hence CaCO3 moles required = 0.0129825/2 = 0.00649125

Hence mass of CaCO3 in total table = 0.00649125*100 = 0.649125gms

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