November 30, 2015

Homework Help: Chemsitry

Posted by Brit on Tuesday, November 13, 2012 at 6:12pm.

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)

Answer this Question

First Name:
School Subject:

Related Questions

More Related Questions