Chapter 7, Problem 7.60E

Stomach acid is essentially $0.10\text{M}$ $\text{HCl}$. An active ingredient found in a number of popular antacids is calcium carbonate, ${\text{CaCO}}_{\text{3}}$. Calculate the number of grams of ${\text{CaCO}}_{\text{3}}$ needed to exactly react with $250.\text{mL}$ of stomach acid.

${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+2\text{HCl}\left(\text{aq}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right)+{\text{CaCl}}_2\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$