Question

What volume in ml of 0.8 M HCl is needed to completely react with 91 grams of the CaCO₃?

Answer 1

2HCl+CaCO₃=CaCl₂+H₂O+CO₂

n (CaCO₃)=$m(CaCO3)/M(CaCO3)$ $=$ $91 g/100 g/mol$ = 0.91 mol

n(HCl)=$2*n(CaCO3)$ $=$ $2*0.91 mol= 1.82 mol$

V(HCl)=$n(HCl)/c(HCl)$ = $1.82 mol/ 0.8 mol/L$ = 2.275 L= 2275 ml

Answer: 2275 ml