Answer
$m(CaCO_{3}) =150mg$
Work Step by Step
The reaction of neutralization of calcium - carbonate by hydrochloric acid is as following:
$CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}$
The amount of $CaCO_{3}$ is two times less than the amount of hydrochloric acid, considering stoichiometric coefficients. Therefore,
$n(CaCO_{3}) = \frac{1}{2}n(HCl) = \frac{1}{2}c(HCl)\times V(HCl)=\frac{1}{2}\times 0.112 \frac{mol}{dm^3}\times 0.0268 dm^3 = 0.0015mol$
This amount can be easily converted to the mass of calcium - carbonate contained in one tablet:
$m(CaCO_{3}) = M(CaCO_{3})\times n(CaCO_{3}) = 100 \frac{g}{mol} \times 0.0015mol = 0.15g = 150mg$
