2.68 g of a mixture of acetaldehyde and glucose were dissolved in water and the resulting solution was

2.68 g of a mixture of acetaldehyde and glucose were dissolved in water and the resulting solution was added to an ammoniacal silver oxide solution prepared from 35.87 ml of a 34% silver nitrate solution (density 1.4 g / ml). The precipitate that formed was filtered off, and an excess of potassium chloride solution was added to the filter, neutralized with nitric acid. In this case, 5.74 g of precipitate fell out. Determine the percentage of the original mixture.

Given:
m mixture (CH3COH, C6H12O6) = 2.68 g
V solution (AgNO3) = 35.87 ml
ω (AgNO3) = 34%
ρ solution (AgNO3) = 1.4 g / ml
m (sediment 2) = 5.74 g

Find:
ω (CH3COH) -?
ω (C6H12O6) -?

Solution:
1) AgNO3 + KOH + 2NH3 => [Ag (NH3) 2] OH + KNO3;
CH3COH + 2 [Ag (NH3) 2] OH => CH3COOH + 2Ag ↓ + 4NH3 + 2H2O;
C6H12O6 + 2 [Ag (NH3) 2] OH => C6H12O7 + 2Ag ↓ + 4NH3 + 2H2O;
[Ag (NH3) 2] OH + HNO3 => AgNO3 + KOH + 2NH3;
AgNO3 + KCl => AgCl ↓ + KNO3;
2) m solution (AgNO3) = ρ solution (AgNO3) * V solution (AgNO3) = 1.4 * 35.87 = 50.218 g;
3) m (AgNO3) = ω (AgNO3) * m solution (AgNO3) / 100% = 34% * 50.218 / 100% = 17.074 g;
4) n (AgNO3) = m (AgNO3) / M (AgNO3) = 17.074 / 170 = 0.1 mol;
5) n total ([Ag (NH3) 2] OH) = n (AgNO3) = 0.1 mol;
6) n (AgCl) = m (AgCl) / M (AgCl) = 5.74 / 143.5 = 0.04 mol;
7) n (AgNO3) = n (AgCl) = 0.04 mol;
8) n rest. ([Ag (NH3) 2] OH) = n (AgNO3) = 0.04 mol;
9) n react. ([Ag (NH3) 2] OH) = n total. ([Ag (NH3) 2] OH) – n rest. ([Ag (NH3) 2] OH) = 0.1 – 0.04 = 0.06 mol;
10) Let n (CH3COH) = (x) r; and n (C6H12O6) = (y) d;
11) n react. 1 ([Ag (NH3) 2] OH) = n (CH3COH) * 2 = (2x) mol;
12) n react. 2 ([Ag (NH3) 2] OH) = n (C6H12O6) * 2 = (2y) mol;
13) m (CH3COH) = n (CH3COH) * M (CH3COH) = (x * 44) g;
14) m (C6H12O6) = n (C6H12O6) * M (C6H12O6) = (y * 180) g;
44x + 180y = 2.68;
2x + 2y = 0.06;
y = 0.03 – x;
44x + 180 * (0.03 – x) = 2.68;
x = 0.02;
15) m (CH3COH) = x * 44 = 0.02 * 44 = 0.88 g;
16) ω (CH3COH) = m (CH3COH) * 100% / m mixture (CH3COH, C6H12O6) = 0.88 * 100% / 2.68 = 32.84%;
17) ω (C6H12O6) = 100% – ω (CH3COH) = 100% – 32.84% = 67.16%.

Answer: The mass fraction of CH3COH is 32.84%; C6H12O6 – 67.16%.