Question

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction,

CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (l)

What mass of CaCO₃ is required to completely with 25 mL of 0.75 MHCl?

Answer 1

CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (l)

1 mol                     2 mol

(40 + 12 + 48) g   2(1 + 35.5) g

= 100 g                 = 73 g

Amount of HCl in 25 mL of 0.75 M HCl

= M x V = 0.75 mpl L^-1 x 25 mL

= 0.75 mol L^-1 x 25/1000 L = {0.75 x 25}/{1000} mol

= 0.0188 mol

2 mol of HCl reacts with 100 g CaCO₃

0.0188 mol of HCl reacts with = {100 g x 0.0188}/{2} g CaCO₃

= 0.94 g CaCO₃