1. A 1.37 grams antacid tablet is added to 25.00 mL of 0.450 M HCl. After the re

Question

1. A 1.37 grams antacid tablet is added to 25.00 mL of 0.450 M HCl. After the resulting reaction was complete, back titration required 10.4 mL of 0.1212 M NaOH. Assuming that the tablet contained only magnesium carbonate and inert binder, calculate the mass percent of magnesium carbonate in the tablet.
2. If a tablet contains 110 mg of Mg(OH)2 and 550 mg of CaCO3, how many moles of HCl would theoretically be needed to neutralized it?
3. If you use 0.20g of an antacid tablet that weighs 1.45g, what percentage of the tablet have you used? If the tablet contains 250mg CaCO3, how much CaCO3 do you have in your measured portion?
Hi can somone help me with these questions? Thank you in advance.

Explanation / Answer

1. MgCO3 reacts with HCl according to the balanced equation:

MgCO3 + 2 HCl = MgCl2 + H2O + CO2

After the reaction is complete, exess HCl is back titrated by NaOH.

25.00 mL of 0.450 M HCl has been added to the antacid tablet. 25.00 mL of 0.450 M HCl contains (25.00 * 0.450/ 1000) mole HCl = 0.011225 mole HCl.

10.4 mL of 0.1212 M NaOH has been used for back titration. Number of moles of NaOH used = (10.4 * 0.1212/1000) mole = 0.00126 mole

0.00126 mole naOH neutralises 0.00126 mole HCl. So, out of 0.01125 mole HCl, 0.00126 mole remains unreacted whereas (0.01125-0.00126) mole = 0.00999 mole HCl has reacted with MgCO3.

From the reaction stoichiometry , we find:

Number of moles of MgCO3 reacted = (1/2) * Number of moles of HCl reacted = (1/2) * 0.00999 mole = 0.004995 mole.

Mass of MgCO3 reacted = Number of moles * Molar mass of MgCO3 = 0.004995 mole * 84 g/ mole = 0.420 g

1.37 grams antacid tablet contains 0.420 g MgCO3

Mass percentage of MgCO3 = (0.420 g / 1.37 g) *100% = 30.65%

2.

Molar mass of Mg(OH)2 is 58 g/ mole. Moles of Mg(OH)2 present in 110 mg (or 0.11 grams) = (mass/ molar mass) = (0.11 grams/ 58 g/ mole) = 0.0019 mole

Mg(OH)2 reacts with HCl according to the balanced equation:

Mg(OH)2+ 2 HCl = MgCl2 + 2 H2O

From the reaction stoichiometry , we find:

Number of moles of HCl reacted = 2* Number of moles of Mg(OH)2 reacted = 2 * 0.0019 mole = 0.0038 mole

Molar mass of CaCO3 is 100 g/ mole. Moles of CaCO3 present in 550 mg (or 0.55 grams) = (mass/ molar mass) = (0.55 grams/ 100 g/ mole) = 0.0055 mole

CaCO3 reacts with HCl according to the balanced equation:
CaCO3 + 2 HCl = CaCl2 + H2O + CO2

From the reaction stoichiometry , we find:

Number of moles of HCl reacted = 2* Number of moles of CaCO3 reacted = 2 * 0.0055 mole = 0.011 mole

Total moles of HCl reacted with Mg(OH)2 and CaCO3 = (0.0038 + 0.011) mole = 0.0148 mole

So, 0.0148 moles of HCl is needed to neutralise the tablet.

3.

Percent of used tablet = (0.20g/ 1.45 g) *100% = 13.8%

Amount of CaCO3 in the used portion = 250 mg * ( 0.20g / 1.45 g) = 34.48 mg

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