The active ingredient in an antacid tablet is calcium carbonate, CaCO_3. One tab

Question

The active ingredient in an antacid tablet is calcium carbonate, CaCO_3. One tablet of a generic brand of antacid contains 270 mg of CaCO_3 What is the molar mass of CaCO_3? How many moles of CaCO_3 are in one roll, 12 tablets, of antacid? If a person takes two tablets, how many grams of calcium are obtained? If the daily recommended quantity of Ca^2+ to maintain bone strength m o women is 1500 mg, how many tablets are needed each day to supply the needed. If two tablets are dissolved in 250 ml of water, what is the [Ca^2+] of the solution? Calcium carbonate reacts with aqueous HCl to make CO_2(g), H_2 CaC l (aq). Write a balanced equation for this reaction. If two tablets of antacid are added to 20.00 mL of 0.400 M HCl, how grams of CO_2 are produced? What is the volume of CO_2 produced for this reaction if P = 1.00 atm and T = 37 degree C?

Explanation / Answer

According to problem one tablet contains 270 mh CaCO3 so two tablets have 540 mg or 0.54 g CaCO3.

Number of moles of CaCO3 = 0.54 g / molar mass, 100.09 g/ moles

= 0.0054 moles CaCO3

CaCO3 = Ca2+   + CO32-

One mole CaCO3 = 1 mole Ca2+

0.0054 moles CaCO3=0.0054 moles Ca2+

Volume = 250 ml = 0.250 L

Concentration = number of moles / volume in L

= 0.0054/ 0.250

= 0.022 M Ca2+

g)

2HCl + CaCO3 ------> CaCl2 + H2O + CO2

Moles of HCl = molarity * volume in L

= 0.400*0.020

= 0.008 moles HCl

Now moles of CO2 = 0.008 moles HCl * 1 mole CO2/2 mole HCl

= 0.004 moles CO2

Amount of CO2 = 0.004 moles CO2*44.01 g/ mole

= 0.176 g

h)

PV = n RT

Here n = 0.004 moles CO2

T = 37 = 310 K

P = 1.0 atm

R = 0.08206 L –atm / mole-K

volume , V = n RT / P

= 0.004*0.08206*310/1.00

= 0.102 L

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