During the heat treatment of copper (II) nitrate weighing 94 g, part of the substance decomposed and 11.2 liters of a mixture

During the heat treatment of copper (II) nitrate weighing 94 g, part of the substance decomposed and 11.2 liters of a mixture of gases were released. To the resulting solid residue was added 292 g of a 10% hydrochloric acid solution. Determine the mass fraction of hydrochloric acid in the resulting solution.

Let us write the equation for the thermal decomposition of copper (II) nitrate:
2Cu (NO3) 2 → 2CuО + 4NO2 + O2 + {Cu (NO3) 2} rest. (1),
where {Cu (NO3) 2} rest. – undecomposed part of copper (II) nitrate.
Thus, the solid residue is a mixture of the formed copper (II) oxide and the remaining copper (II) nitrate.
Only one component of the solid residue reacts with hydrochloric acid – the formed CuO:
CuO + 2HCl → CuCl2 + H2O (2)
Let’s calculate the amount of substance of the resulting gas mixture:
n (NO2 + O2) = 11.2 L / 22.4 L / mol = 0.5 mol.
From equation (1): n (CuO) = n (NO2 + O2) ∙ 2/5 = 0.5 mol ∙ 2/5 = 0.2 mol.
Using equation (2), we calculate the amount of hydrochloric acid substance that has reacted with CuO:
n (HCl (reac.)) = 2 ∙ n (CuO) = 2 ∙ 0.2 mol = 0.4 mol.
Let’s find the total mass and amount of the hydrochloric acid substance taken for the reaction:
m (HCl (gen.)) substances = m (HCl (gen.)) solution ∙ ω (HCl) = 292 g ∙ 0.1 = 29.2 g
n (HCl (gen.)) = m (HCl (gen.)) in islands / M (HCl) = 29.2 g / 36.5 g / mol = 0.8 mol.
Let’s find the amount of substance and the mass of the remaining hydrochloric acid in the resulting solution:
n (HCl (rest)) = n (HCl (total)) – n (HCl (reaction)) = 0.8 mol – 0.4 mol = 0.4 mol.
m (HCl (rest)) = n (HCl (rest)) ∙ M (HCl) = 0.4 mol ∙ 36.5 g / mol = 14.6 g.
Let us find the mass of the final solution mfin.r-ra:
mkon.r-pa = m (CuO) + m (Cu (NO3) 2 (rest)) + m (HCl (total)) solution
Let’s calculate the mass of the formed СuO:
m (CuO) = n (CuO) ∙ M (CuO) = 0.2 mol ∙ 80 g / mol = 16 g.
Let us calculate the mass of undecomposed Cu (NO3) 2:
n (Cu (NO3) 2 (reac.)) = n (CuO) = 0.2 mol,
where Cu (NO3) 2 (reactive) is the decomposed part of copper (II) nitrate.
m (Cu (NO3) 2 (reactive)) = n (Cu (NO3) 2 (reactive)) ∙ M (Cu (NO3) 2) = 0.2 mol ∙ 188 g / mol = 37.6 g
m (Cu (NO3) 2 (rest)) = m (Cu (NO3) 2 (initial)) – m (Cu (NO3) 2 (reactive)) = 94 g – 37.6 g = 56.4 g.
mcone solution = m (CuO) + m (Cu (NO3) 2 (rest)) + m (HCl (total)) solution = 16g + 56.4g + 292g = 364.4g
Let us determine the mass fraction of hydrochloric acid in the resulting solution ω (HCl) end solution:
ω (HCl) end solution = m (HCl (rest)) / m end solution = 14.6g / 364, 4g = 0.0401 (4.01%)
Answer: ω (HCl) = 4.01%

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