Question

5. Calculate how many grams of HCl can be neutralized by: a. 150 mg of CaCO3 b. 150 mg of Mg(OH)2 c. 150 mg of NaHCO3 On a weight basis only, which would be the most effective?

6. Osteoporosis is a disease that strikes older women resulting in a loss of calcium from the bone. To prevent this disease, it is recommended that older women consume from 1,200 to 1,500 mg of calcium per day. A woman said, “I'm taking 1,000 mg of calcium a day because I take two TumsTM”. How much calcium was she actually ingesting? (Tums contains 500 mg CaCO3 /tablet.)   Antacid titration

7. Calculate how many grams of HCl are present in 250 mL of 0.2016 N HCl. 016 moles are present in 1litre 250 ml will contain = 0.2016/4 = 0.0504 moles Now mass = mole * molecular mass 0.0504*36.46 = 1.8375 grams

. 8. One package of an antacid that was labeled REGULAR contained 150 tablets. Each tablet consisted of 500 mg of CaCO3. A second package of an antacid was labeled EXTRA STRENGTH and it contained 96 tablets. Each of these tablets consisted of 700 mg of CaCO3. Both packages sold for $3.83. Which is the better value?

Answer 1

5.

(a)

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

from the balanced equation,

100 g. of CaCO3 can neutralise 2 * 36.5 = 73 g. of HCl

Then,

150 mg or 0.150 g. of CaCO3 can neutralise 0.150 * 73 / 100 = 0.110 g. of HCl = 110 mg. of HCl

(b)

2 HCl (aq.) + Mg(OH)2 (aq.) -------------> MgCl2 (aq.) + 2 H2O (l)

From the balanced equation,

58 g. of Mg(OH)2 can neutralise 73 g. of HCl

Then,

0.150 g. of Mg(OH)2 can neutralise 0.150 * 73 / 58 = 0.189 g. of HCl = 189 mg. of HCl

(c)

HCl (aq.) + NaHCO3 (aq.) -------------> NaCl (aq.) + H2O (l) + CO2 (g)

From the balanced equation,

84 g. of NaHCO3 can neutrlaise 36.5 g. of HCl

Then,

0.150 g. of NaHCO3 can neutralise 0.150 * 36.5 / 84 = 0.0652 g. of HC= 65.2 mg. of HCl

SO, on weight basis Mg(OH)2 is most effective to neutralise HCl.