Question

Calcium carbonate reacts with hydrochloric acid according to the equation:

CaCO_3(aq)  +  2 HCl_(aq)  -->  CO_2(g)  +  CaCl_2(aq)  +  H₂O_(l)

What volume in ml of 1.1 M HCl is needed to completely react with 67 grams of the CaCO₃?

Answer 1

Moles CaCO₃ $=\frac{67}{100}=0.67 moles$

Moles HCl $=0.67×2=1.34moles$

Volume in ml = $\frac{1.34×1000}{1.1}=1218.18ml$