JEE Challenger
More from d and f Block Elements
ChemHardAdv 2022 • Paper 1

Mass Calculation of Precipitate X from Copper Nitrate Reaction

The treatment of an aqueous solution of 3.74 g3.74\text{ g} of Cu(NO3)2\text{Cu(NO}_3\text{)}_2 with excess KI\text{KI} results in a brown solution along with the formation of a precipitate. Passing H2S\text{H}_2\text{S} through this brown solution gives another precipitate X\mathbf{X}. The amount of X\mathbf{X} (in g\text{g}) is ______.

[Given: Atomic mass of H=1,N=14,O=16,S=32,K=39,Cu=63,I=127\text{H} = 1, \text{N} = 14, \text{O} = 16, \text{S} = 32, \text{K} = 39, \text{Cu} = 63, \text{I} = 127]

Official Numerical Answer0.32

Topics & Concepts

d and f Block ElementsStoichiometryMole ConceptRedox Reactions

Step-by-Step Solution

To calculate the amount of precipitate X\mathbf{X} formed, we break down the chemical processes step-by-step:

Step 1: Determination of moles of Cu(NO3)2\text{Cu(NO}_3)_2

The molar mass of copper(II) nitrate, Cu(NO3)2\text{Cu(NO}_3)_2, is calculated as: Molar mass of Cu(NO3)2=63+2×(14+3×16)=63+2×62=187 g/mol\text{Molar mass of Cu(NO}_3)_2 = 63 + 2 \times (14 + 3 \times 16) = 63 + 2 \times 62 = 187 \text{ g/mol}

The number of moles of Cu(NO3)2\text{Cu(NO}_3)_2 present in 3.74 g3.74\text{ g} is: nCu(NO3)2=3.74 g187 g/mol=0.02 moln_{\text{Cu(NO}_3)_2} = \frac{3.74\text{ g}}{187\text{ g/mol}} = 0.02\text{ mol}

Step 2: Reaction of Cu(NO3)2\text{Cu(NO}_3)_2 with excess KI\text{KI}

When Cu(NO3)2\text{Cu(NO}_3)_2 reacts with excess KI\text{KI}, Cu2+\text{Cu}^{2+} ions oxidize I−\text{I}^- to iodine (I2\text{I}_2) and are themselves reduced to cuprous iodide (CuI\text{CuI} or Cu2I2\text{Cu}_2\text{I}_2), which precipitates out: 2Cu(NO3)2+4KI→Cu2I2↓+I2+4KNO32\text{Cu(NO}_3)_2 + 4\text{KI} \rightarrow \text{Cu}_2\text{I}_2\downarrow + \text{I}_2 + 4\text{KNO}_3

The liberated iodine (I2\text{I}_2) dissolves in excess KI\text{KI} solution to form KI3\text{KI}_3, giving a characteristic brown color to the solution.

From the stoichiometry of the reaction: nI2=12×nCu(NO3)2=12×0.02 mol=0.01 moln_{\text{I}_2} = \frac{1}{2} \times n_{\text{Cu(NO}_3)_2} = \frac{1}{2} \times 0.02\text{ mol} = 0.01\text{ mol}

Step 3: Reaction of the brown solution with H2S\text{H}_2\text{S}

When hydrogen sulfide (H2S\text{H}_2\text{S}) gas is passed through the brown solution containing dissolved iodine (I2\text{I}_2), I2\text{I}_2 acts as an oxidizing agent and oxidizes H2S\text{H}_2\text{S} to elemental sulfur (S\text{S}), which forms precipitate X\mathbf{X}: I2+H2S→2HI+S↓(X)\text{I}_2 + \text{H}_2\text{S} \rightarrow 2\text{HI} + \text{S}\downarrow (\mathbf{X})

From the stoichiometry: nS=nI2=0.01 moln_{\text{S}} = n_{\text{I}_2} = 0.01\text{ mol}

Step 4: Calculation of the mass of precipitate X\mathbf{X}

The atomic mass of sulfur (S\text{S}) is 32 g/mol32\text{ g/mol}. Mass of X=nS×Molar mass of S=0.01 mol×32 g/mol=0.32 g\text{Mass of } \mathbf{X} = n_{\text{S}} \times \text{Molar mass of S} = 0.01\text{ mol} \times 32\text{ g/mol} = 0.32\text{ g}

Final Answer: The amount of precipitate X\mathbf{X} is 0.32 g.

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