posted by Brit on .
Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of grams of CaCO3 needed to exactly react with 250 mL of stomach acid.
CaCO3(s)+ 2HCl(aq)= CO2(g)+ CaCl2(aq)+ H20(l)
I answered this below.