Through 160 g of a potassium hydroxide solution, carbon monoxide (IV) with a volume of 17.92

Through 160 g of a potassium hydroxide solution, carbon monoxide (IV) with a volume of 17.92 liters was passed until a medium salt was formed. Determine the mass fraction of cadium in the initial solution.

Given:
m solution (KOH) = 160 g
V (CO2) = 17.92 l
Vm = 22.4 l / mol

Find:
ω (KOH) -?

Solution:
1) 2KOH + CO2 => K2CO3 + H2O;
2) M (KOH) = Mr (KOH) = Ar (K) * N (K) + Ar (O) * N (O) + Ar (H) * N (H) = 39 * 1 + 16 * 1 + 1 * 1 = 56 g / mol;
3) n (CO2) = V (CO2) / Vm = 17.92 / 22.4 = 0.8 mol;
4) n (KOH) = n (CO2) * 2 = 0.8 * 2 = 1.6 mol;
5) m (KOH) = n (KOH) * M (KOH) = 1.6 * 56 = 89.6 g;
6) ω (KOH) = m (KOH) * 100% / m solution (KOH) = 89.6 * 100% / 160 = 56%.

Answer: Mass fraction of KOH is 56%.