The gas obtained under the action of hydrochloric acid ha of calcium carbonate weighing 25 grams

The gas obtained under the action of hydrochloric acid ha of calcium carbonate weighing 25 grams was absorbed with a solution of sodium hydroxide (thus forming a medium salt). Calculate the volume of sodium hydroxide (mass fraction of NaOH 8%), density 10.9 g / ml, which is required for absorption of the formed gas.

Let’s write down the reaction scheme and arrange the coefficients:
CaCO3 + 2HCl = CaCl2 + H2O + CO2
Let us determine the amount of substance n (CaCO3) = m (CaCO3) / M (CaCO3), where m (CaCO3) is the mass of calcium carbonate, M (CaCO3) = 100 g / mol is the molar mass of calcium carbonate.
n (CaCO3) = 25/100 = 0.25 mol. The reaction equation shows n (CaCO3) = n (CO2) – the amount of substance CO2
Let’s write the following reaction:
CO2 + 2NaOH = Na2CO3 + H2O
Using this reaction equation, we determine the amount of NaOH substance – n (NaOH) = n (CO2) = 0.25 mol.
Let us find the mass of the solute NaOH m (r.v. NaOH) = n (NaOH) * m (NaOH), where M (NaOH) = 40 g / mol is the molar mass of NaOH.
m (r.v. NaOH) = 0.25 * 40 = 10 g
Let us find the mass of the NaOH solution m (p-pa NaOH) = m (r.v. NaOH) / w, where w is the mass fraction of NaOH in the solution
m (p-pa NaOH) = 10 / 0.08 = 125 g
Determine the volume of the NaOH solution V (NaOH) = m (p-pa NaOH) / p, where p is the density of the solution
V (NaOH) = 125 / 10.9 = 11.47 ml