The gas obtained by the action of an excess of hydrochloric acid on sodium carbonate weighing 10.6 g

The gas obtained by the action of an excess of hydrochloric acid on sodium carbonate weighing 10.6 g was passed through a solution weighing 50 g with a mass fraction of potassium hydroxide 22.4%. determine the mass fraction of salt in the resulting solution.

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

Find the amount of sodium carbonate substance:

n (Na2CO3) = m (Na2CO3) / M (Na2CO3) = 10.6 / 106 = 0.1 mol;

According to the stoichiometry of the reaction:

n (CO2) = n (Na2CO3) = 0.1 mol;

m (CO2) = n (CO2) * M (CO2) = 0.1 * 44 = 4.4 g;

Find the mass and amount of the substance of potassium hydroxide:

m (KOH) = w (KOH) * m (KOH solution) = 0.224 * 50 = 11.2 g;

n (KOH) = m (KOH) / M (KOH) = 0.2 mol;

The amount of alkali is sufficient for the reaction of formation of potassium carbonate (and not bicarbonate) to proceed:

CO2 + 2KON = K2CO3 + H2O

n (K2CO3) = n (CO2) = 0.1 mol;

Let’s find the mass of potassium carbonate and its mass fraction in the solution:

m (K2CO3) = n (K2CO3) * M (K2CO3) = 0.1 * 138 = 13.8 g;

m (K2CO3 solution) = m (KOH solution) + m (CO2) = 50 + 4.4 = 54.4 g;

w (K2CO3) = m (K2CO3) / m (K2CO3 solution) = 13.8 / 54.4 = 0.2537 = 25.37%.

Answer: w (K2CO3) = 25.37%.