The chemical equation representing the reaction of calcium carbonate with hydrochoric acid is:
[tex]CaCO_{3}(s)+2HCl(aq)-->CaCl_{2}(aq)+H_{2}O(l)+CO_{2}( g)[/tex]
Moles of HCl given = [tex]25.0mL*\frac{1L}{1000mL}* \frac{0.10mol}{L} =0.0025molHCl[/tex]
Moles of [tex]CaCO_{3}[/tex]=[tex]0.055 g *\frac{1mol}{100.09g} =5.50*10^{-4}molCaCO_{3}[/tex]
Moles of HCl that would react with [tex]5.50*10^{-4}molCaCO_{3}[/tex]:
[tex]5.50*10^{-4}molCaCO_{3}* \frac{2mol HCl}{1 mol CaCO_{3} }[/tex]=0.0011mol HCl
Moles of HCl that would not be neutralized = 0.0025 mol - 0.0011 mol
= 0.0014 mol HCl
