To obtain aluminum hydroxide, 200 ml (density 1.07 g / ml) of aluminum nitrate solution and 400 ml

To obtain aluminum hydroxide, 200 ml (density 1.07 g / ml) of aluminum nitrate solution and 400 ml (density 1.1 g / ml) of potassium hydroxide solution were taken. This formed 28 g of sediment with a loss of 8%. Determine the% content (mass fraction of substances) in the original solutions.

Given:
V solution (Al (NO3) 3) = 200 ml
ρ solution (Al (NO3) 3) = 1.07 g / ml
V solution (KOH) = 400 ml
ρ solution (KOH) = 1.1 g / ml
m (sediment) = 28 g
η = 92%

To find:
ω (Al (NO3) 3) -?
ω (KOH) -?

Decision:
1) Al (NO3) 3 + 3KOH => Al (OH) 3 ↓ + 3KNO3;
2) n practical (Al (OH) 3) = m (Al (OH) 3) / M (Al (OH) 3) = 28/78 = 0.359 mol;
3) n theory. (Al (OH) 3) = n practical. (Al (OH) 3) * 100% / η = 0.359 * 100% / 92% = 0.39 mol;
4) n (Al (NO3) 3) = n theory. (Al (OH) 3) = 0.39 mol;
n (KOH) = n theory. (Al (OH) 3) * 3 = 0.39 * 3 = 1.17 mol;
5) m (Al (NO3) 3) = n (Al (NO3) 3) * M (Al (NO3) 3) = 0.39 * 213 = 83.07 g;
m (KOH) = n (KOH) * M (KOH) = 1.17 * 56 = 65.52 g;
6) m solution (Al (NO3) 3) = ρ solution (Al (NO3) 3) * V solution (Al (NO3) 3) = 1.07 * 200 = 214 g;
m solution (KOH) = ρ solution (KOH) * V solution (KOH) = 1.1 * 400 = 440 g;
7) ω (Al (NO3) 3) = m (Al (NO3) 3) * 100% / m solution (Al (NO3) 3) = 83.07 * 100% / 214 = 38.82%;
ω (KOH) = m (KOH) * 100% / m solution (KOH) = 65.52 * 100% / 440 = 14.89%.

Answer: The mass fraction of Al (NO3) 3 is 38.82%; KOH – 14.89%.