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

Carbon monoxide was passed through a solution with a volume of 128 cm3 (q = 1.09 g / cm3) with a mass fraction of potassium hydroxide of 10% 4. Determine the mass of the formed average salt

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

Find:
m (salt) -?

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

Answer: The mass of K2CO3 is 17.25 g.