When a sample of silver nitrate was heated, part of the substance decomposed and a gas mixture with a volume of 6.72 liters (in terms of NU) was released. The mass of the residue was 25 g. After that, the residue was placed in 50 ml of water and 18.25 g of 20% hydrochloric acid solution were added. Determine the mass fraction of hydrochloric acid in the resulting solution.
Let us write the equation for the thermal decomposition of silver nitrate (I):
2AgNО3 → 2Ag + 2NО2 + O2 (1)
The solid residue is a mixture of the resulting silver and the remaining silver (I) nitrate.
Only silver (I) nitrate reacts with hydrochloric acid:
AgNO3 + HCl → AgCl ↓ + HNO3 (2)
Let’s calculate the amount of gases formed during the decomposition of silver nitrate:
n (NO2 + O2) = 6.72 L / 22.4 L / mol = 0.3 mol.
According to equation (1) n (Ag) = 2/3 ∙ n (NO2 + O2) = 2/3 ∙ 0.3 mol = 0.2 mol
m (AgNO3) oc. = 25 g – 21.6 g = 3.4 g
n (AgNO3) oc. = 3.4 g / 170 g / mol = 0.02 mol.
We calculate the mass and amount of the substance of hydrochloric acid in its initial solution:
m (HCl) and cx. = 18.25 g ∙ 0.2 = 3.65 g
n (HCl) and cx. = 3.65 g / 36.5 g / mol = 0.1 mol
According to equation (2) n (AgNO3) oc. = N (AgCl) = n (HCl) reactive, where n (HCl) reactive. – the amount of hydrochloric acid substance that went into the reaction with AgNO3. Therefore, the amount of the substance and the mass of unreacted hydrochloric acid:
n (HCl) rest = 0.1 mol – 0.02 mol = 0.08 mol;
m (HCl) rest. = 0.08 mol ∙ 36.5 g / mol = 2.92 g.
Let’s calculate the mass of the precipitate
m (AgCl) = n (AgCl) ∙ M (AgCl) = 0.02 mol ∙ 143.5 g / mol = 2.87 g.
The mass of the resulting solution is:
mcon.p-pa = residual + m (HCl) solution + m (H2O) – m (AgCl) = 3.4 g + 18.25 g + 50 g – 2.87 g = 68.78 g.
The mass fraction in the resulting hydrochloric acid solution is equal to:
ω (НСl) = m (НСl) rest./mcon.p-pa = 2.92 g / 68.78 g = 0.0425 (4.25%).
Answer: ω (НСl) = 4.25%.
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