# 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). Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.