Chlorine reacted without residue with 445 ml of hot 50% potassium hydroxide solution (density 1.51 g / ml).
Chlorine reacted without residue with 445 ml of hot 50% potassium hydroxide solution (density 1.51 g / ml). Determine the mass fractions of substances in the resulting solution.
Given:
V solution (KOH) = 445 ml
ω (KOH) = 50%
ρ solution (KOH) = 1.51 g / ml
Find:
ω (in-in) -?
Solution:
1) 3Cl2 + 6KOH => KClO3 + 5KCl + 3H2O;
2) m solution (KOH) = ρ solution (KOH) * V solution (KOH) = 1.51 * 445 = 671.95 g;
3) m (KOH) = ω (KOH) * m solution (KOH) / 100% = 50% * 671.95 / 100% = 335.975 g;
4) n (KOH) = m (KOH) / M (KOH) = 335.975 / 56 = 6 mol;
5) n (KClO3) = n (KOH) / 6 = 6/6 = 1 mol;
6) m (KClO3) = n (KClO3) * M (KClO3) = 1 * 122.5 = 122.5 g;
7) n (KCl) = n (KOH) * 5/6 = 6 * 5/6 = 5 mol;
8) m (KCl) = n (KCl) * M (KCl) = 5 * 74.5 = 372.5 g;
9) n (Cl2) = n (KOH) * 3/6 = 6 * 3/6 = 3 mol;
10) m (Cl2) = n (Cl2) * M (Cl2) = 3 * 71 = 213 g;
11) m solution = m (Cl2) + m solution (KOH) = 213 + 671.95 = 884.95 g;
12) ω (KClO3) = m (KClO3) * 100% / m solution = 122.5 * 100% / 884.95 = 13.84%;
13) ω (KCl) = m (KCl) * 100% / m solution = 372.5 * 100% / 884.95 = 42.09%.
Answer: The mass fraction of KClO3 is 13.84%; KCl – 42.09%.
