Carbon monoxide IV was passed through a solution with a volume of 128 cm3 (p = 1.09 g / cm3)

Carbon monoxide IV was passed through a solution with a volume of 128 cm3 (p = 1.09 g / cm3) with a mass fraction of potassium hydroxide of 10%. Determine the mass of the medium salt formed

Given:
V solution (KOH) = 128 cm3 = 128 ml
ρ solution (KOH) = 1.09 g / cm3 = 1.09 g / ml
ω (KOH) = 10%

To find:
m (salt) -?

Solution:
1) 2KOH + CO2 => K2CO3 + H2O;
2) m solution (KOH) = ρ solution (KOH) * V solution (KOH) = 1.09 * 128 = 139.52 g;
3) m (KOH) = ω (KOH) * m solution (KOH) / 100% = 10% * 139.52 / 100% = 13.952 g;
4) n (KOH) = m (KOH) / M (KOH) = 13.952 / 56 = 0.249 mol;
5) n (K2CO3) = n (KOH) / 2 = 0.249 / 2 = 0.125 mol;
6) m (K2CO3) = n (K2CO3) * M (K2CO3) = 0.125 * 138 = 17.25 g.

Answer: The mass of K2CO3 is 17.25 g.