Through 100 g of a 5.6% potassium hydroxide solution, 22.4 liters of carbon monoxide (IV) were passed.

Through 100 g of a 5.6% potassium hydroxide solution, 22.4 liters of carbon monoxide (IV) were passed. Determine the mass fraction of the formed salt in the solution.

2KOH + CO2 → K2CO3 + H2O
m (KOH) = 100 x 0.056 = 5.6 g
n (KOH) = 5.6 / 56 = 0.1 mol – deficient
n (CO2) = 22.4 / 22/4 = 1 mol – in excess
n (K2CO3) = 0.05 mol
m (K2CO3) = 0.05 x (39 x 2 + 14 +48) = 7 g
m (solution) = 100 + 1 x (12 + 32) = 100 + 44 = 144 g
ω (K2CO3) = 7/144 x 100% = 4.8%
Answer: the mass fraction of the formed salt in the solution (potassium carbonate) is 4.8%.