Determine what salt is formed during the passage of CO₂ weighing 0.88 grams

Determine what salt is formed during the passage of CO₂ weighing 0.88 grams through a solution of sodium hydroxide weighing 5 grams, with a mass fraction of the meadow of 20%. Calculate the mass of this salt.

There are two possible variants of the reaction: the formation of bicarbonate and carbonate with an excess of alkali. CO2 + H2O H2CO3. H2CO3 + NaOH -> NaHCO3 + H2O. H2CO3 + 2NaOH (g) -> Na2CO3 + 2H2O. Let’s calculate the number of moles of substances: n (CO₂) = m / M = 0.88 / 44 = 0.02 mol. m (NaOH) = 5 * 20% / 100% = 1 g. n (NaOH) = m / M = 1/40 = 0.025 mol. The ratio shows that the alkali will be in excess: n (NaOH): n (CO₂) = 0.025: 0.02 = 1.25: 1. This means that at first the reaction will proceed with the formation of an acidic salt NaHCO₃ with the participation of 0.02 mol of alkali, and the remainder of 0.005 mol will react to form bicarbonate. n (Na₂CO₃) = n (NaOH rest) = 0.005 mol, and the remainder 0.02 – 0.005 = 0.015 mol NaHCO₃. m (Na₂CO₃) = 0.005 * 106 = 0.53 g. m (NaHCO₃) = 0.015 * 84 = 1.26 g.