Question

Many common antacid products contain CaCO3, which behaves as a weak base. Each mole of CaCO3 can neutralize 2 moles of HBr according to the reaction below. An antacid tablet has a total mass of 1.62 g and requires 108 mL of 0.099 MHBr to reach the equivalence point during a titration. What is the mass percent of CaCO3 (molar mass - 100. g/mol) in the tablet? 2HBr(aq) +CaCO3(s) ---CaBrzlaq)+H2011) + CO2(8) 1.32% 66.1% 132% 53.5% 0.535% 0.330% 33.0% Next • Previous

Answer 1

lets calculate the mol of HBr
volume , V = 108 mL
= 0.108 L


use:
number of mol,
n = Molarity * Volume
= 9.9*10^-2*0.108
= 1.069*10^-2 mol

According to balanced equation
mol of CaCO3 reacted = (1/2)* moles of HBr
= (1/2)*1.069*10^-2
= 5.346*10^-3 mol
This is number of moles of CaCO3

Molar mass of CaCO3 = 100 g/mol

use:
mass of CaCO3,
m = number of mol * molar mass
= 5.346*10^-3 mol * 100 g/mol
= 0.5351 g

Now use:
Mass % CaCO3 = mass of CaCO3 * 100 / MASS OF TABLET
= 0.5351 * 100 / 1.62
= 33.0 %
Answer: 33.0 %