Question

You measured out 1.212g caco3 to create the calcium supplement. how many ml of 1m hcl would you need to completely dissolve the caco3?

Answer 1

Answer: -

24.24 mL

Explanation: -

Mass of CaCO₃ = 1.212 g

Molar mass of CaCO₃ = 40 x 1 +12 x 1 + 16 x 3 = 100 g / mol

Number of moles of CaCO₃ = $\frac{1.212 g}{100 g/ mol}$

= 0.01212 mol

The balanced chemical equation for the reaction that occurs when CaCO₃ is added to HCl is

CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂

From the equation we see

1 mol of CaCO₃ reacts with 2 mol of HCl

0.01212 mol of CaCO₃ react with $\frac{2 mol HCl x 0.01212 mol CaCO₃}{1 mol CaCO₃}$

= 0.02424 mol of HCl

Strength of HCl = 1 M

Volume of HCl required = Number of moles of HCl / strength of HCl

= 0.02424 mol / 1 M

= 0.02424 L

= 0.02424 x 1000 mL

= 24.24 mL