When a sample of silver (I) nitrate was heated, part of the substance decomposed, with the formation of a solid residue weighing 88 g

When a sample of silver (I) nitrate was heated, part of the substance decomposed, with the formation of a solid residue weighing 88 g To this residue was added 200 g of a 20% hydrochloric acid solution, resulting in a solution weighing 205.3 g with a mass fraction of hydrochloric acid 15, 93%. Determine the volume of the gas mixture released during the decomposition of silver nitrate (I).

Let us write the decomposition equation for silver nitrate (I):
2AgNO3 → 2Ag + 2NO2 + O2 + {AgNO3} rest. (1)
where {AgNO3} rest. – undecomposed part of silver nitrate (I).
Thus, 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)
We calculate the mass and amount of the substance of hydrochloric acid in its initial solution:
m (HCl) and cx. = 20g ∙ 0.2 = 40g
n (HCl) and cx. = 40g / 36.5 g / mol = 1.1 mol
We calculate the mass and amount of the hydrochloric acid substance in the resulting solution:
m (HCl) con. = 205.3 g ∙ 0.1593 = 32.7 g
n (HCl) end = 32.7 g / 36.5 g / mol = 0.896 mol (0.9 mol)
Let’s calculate the amount of hydrochloric acid substance that went into the reaction with AgNO3:
n (HCl) reac = 1.1 mol – 0.896 mol = 0.204 mol (0.2 mol)
Let’s find the amount of the substance and the mass of undecomposed silver nitrate:
According to equation (2) n (AgNO3) oc. = N (HCl) reaction. = 0.204 mol. (0.2 mol)
m (AgNO3) oc. = (AgNO3) oc. ∙ M (AgNO3) = 0.204 mol ∙ 170 g / mol = 34.68 g. (34 g)
Let’s find the mass of the resulting silver:
m (Ag) = residual – m ((AgNO3) oc) = 88 g – 34.68 g = 53.32 g (54 g)
n (Ag) = m (Ag) / M (Ag) = 53.32 g / 108 g / mol = 0.494 mol. (0.5 mol)
Let us find the amount of the substance and the volume of the gas mixture formed during the decomposition of silver nitrate:
According to equation (1) n (NO2 + O2) = 3/2 ∙ n (Ag) = 3/2 ∙ 0.494 mol = 0.741 mol (0.75 mol)
V (NO2 + O2) = n (NO2 + O2) ∙ Vm = 0.741 mol ∙ 22.4 l / mol = 16.6 l. (16.8 l).
Answer: V (NO2 + O2) = 16.6 liters. (16.8L).

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