Please help! When a calcium carbonate tablet is ingested, it dissolves by the reaction of stomach acid...?

Question

which contains hydrochloric acid. The unbalanced equation for this reaction is CaCO3(s)+HCl(aq) ----> CaCl2(aq)+ H2O(l)+CO2(g). How many calcium carbonate tablets, each containing 500 mg of the active ingredient are required to react with 18.5 grams of stomach acid containing 49.5% HCl by mass? please explain

Answer 1

Balanced chenical eqn,
CaCO3(s)+2HCl(aq) ----> CaCl2(aq)+ H2O(l)+CO2(g)

Coclusion from the reaction,
2 moles of HCl neutralizes 1 mole of CaCO3

Given data,
tomach acid contains 49.5% HCl by mass.

calculations,
18.5 gm stomach acid contains = 49.5*18.5/100 = 9.1575 gm HCl
Molecular wt of HCl = 35.5 + 1 = 36
=>9.1575 gm HCl = 9.1575/36.5 moles of HCl =0.2509 moles.

=> reqd moles of CaCO3 = 0.2509/2 = 0.12545(refer to "Coclusion from the reaction")
Molecular wt of CaCO3 = 40 + 12 + 16*3 = 100
=>0.12545 mole CaCO3 = 0.12545 * 100 gm CaCO3= 12.545 gm

each tablet contains 500 mg = 0.5 gm CaCO3
=> reqd no of tablets = 12.545/0.5 = 25.09 = nearly 25

answer,
25 tablets will be reqd.
( 25 looks a round figure but the actual ans will be 26 tablet, since 25.09 tabs are reqd for complete neutrilization of the acid)

Answer 2

First balance the equation:

CaCO3 + 2HCl ---> CaCl2 + H20 + CO2

From this you can see that one mole of CaCO3 will react with 2 moles of HCl. Now you need to find the moles of each substance. The molecular mass of CaCO3 is 100 g/mole. Each tablet contains 0.5/100 = 0..005 moles of CaCO3. 18.5 grams of stomach acid contains 0.495*18.5 = 9.158 grams of HCl The molecular mass of HCl is 36.46, so the stomach contains 9.158/38.46 = 0.238 moles of HCl. You need half that amount in moles of CaCO3, or 0.119 moles of CaCO3. Each tablet contains 0.005 moles, so you need 0.119/0.005 = 23.81 tablets