The treatment of an aqueous solution of 3.74 g of Cu(NO3)2 with excess KI results in a brown solutio

Q: Why does the solution turn brown instead of just forming a precipitate?

The reaction produces $\text{I}_2$, which is normally violet/black, but in the presence of excess $\text{KI}$, it forms the highly soluble triiodide ion ($\text{I}_3^-$), which is deep brown in aqueous solution.

Q: What is the first precipitate formed?

The first precipitate is Cuprous Iodide ($\text{Cu}_2\text{I}_2$ or $\text{CuI}$), which is white. The brown color of the solution often masks it visually, but it is there.

Q: Why doesn't $\text{H}_2\text{S}$ react with the first precipitate?

$\text{Cu}_2\text{I}_2$ is highly insoluble and chemically stable in this environment. $\text{H}_2\text{S}$ prefers to react with the strong oxidizing agent present in the solution, which is the $\text{I}_3^-$ complex.

Q: How do we know $\text{H}_2\text{S}$ oxidizes to Sulfur?

$\text{H}_2\text{S}$ is a classic reducing agent in analytical chemistry. When it reduces halogens (like $\text{I}_2$ or $\text{I}_3^-$), the sulfide ion ($\text{S}^{2-}$) loses electrons to become elemental sulfur ($\text{S}$), which precipitates out.

Q: Can I just use the overall reaction to find the answer?

Yes, if you can correctly combine the two steps! However, it is much safer to go step-by-step to avoid missing stoichiometric coefficients, especially since $\text{KI}$ is in excess and forms an intermediate complex.

Question

The treatment of an aqueous solution of 3.74 g of $	ext{Cu(NO}_3)_2$ with excess KI results in a brown solution along with the formation of a precipitate. Passing $	ext{H}_2	ext{S}$ through this brown solution gives another precipitate X. The amount of X (in g) is ______.
[Given : Atomic mass of H = 1, N = 14, O = 16, S = 32, K = 39, Cu = 63, I = 127]

Sigma Percentile · Accepted Answer

### Unraveling the Mystery of Precipitate X: A Journey Through Redox Stoichiometry

Imagine you are standing in a chemistry lab. You have a beaker filled with a beautiful blue aqueous solution of copper nitrate, $	ext{Cu(NO}_3)_2$. The problem asks us to perform a sequence of reactions and track the exact mass of a final precipitate. This is a classic test of both your knowledge of inorganic reactions and your precision with stoichiometry.

### The Setup: Moles of the Starting Material

Before we dive into the chemical reactions, we must establish our starting point. How much copper nitrate do we actually have?

The problem states we have $3.74 	ext{ g}$ of $	ext{Cu(NO}_3)_2$. To convert this to moles, we first calculate its molar mass.

$$M_{	ext{Cu(NO}_3)_2} = 63 + 2 	imes (14 + 3 	imes 16) = 187 	ext{ g/mol}$$

Now, dividing the given mass by the molar mass gives us the initial moles:

$$n = \frac{3.74}{187} = 0.02 	ext{ mol}$$

We are starting our journey with exactly $0.02 	ext{ moles}$ of copper nitrate.

### The First Reaction: Formation of the Brown Solution

Next, we add excess potassium iodide ($	ext{KI}$) to our beaker. A fascinating redox reaction takes place. The cupric ions ($	ext{Cu}^{2+}$) are reduced to cuprous iodide ($	ext{Cu}_2	ext{I}_2$), which forms a white precipitate. Simultaneously, the iodide ions ($	ext{I}^-$) are oxidized to elemental iodine ($	ext{I}_2$).

However, there is a catch! Because $	ext{KI}$ is in excess, the liberated iodine doesn't just sit there; it reacts with the excess iodide to form the highly soluble triiodide complex ion, $	ext{I}_3^-$. This complex is what gives the resulting solution its deep brown color. The overall balanced equation is:

$$2	ext{Cu(NO}_3)_2 + 5	ext{KI} ightarrow 	ext{Cu}_2	ext{I}_2 \downarrow + 	ext{KI}_3 + 4	ext{KNO}_3$$

Looking at the stoichiometry, $2 	ext{ moles}$ of copper nitrate produce $1 	ext{ mole}$ of $	ext{KI}_3$. Therefore, our $0.02 	ext{ moles}$ of copper nitrate will produce exactly half that amount:

$$n_{	ext{KI}_3} = \frac{0.02}{2} = 0.01 	ext{ mol}$$

### The Second Reaction: Hydrogen Sulfide Enters the Scene

Now, we take this brown solution containing $0.01 	ext{ moles}$ of $	ext{KI}_3$ and pass hydrogen sulfide ($	ext{H}_2	ext{S}$) gas through it. $	ext{H}_2	ext{S}$ is a well-known reducing agent. It reduces the iodine in the triiodide complex back to iodide ions, while it gets oxidized to elemental sulfur ($	ext{S}$).

This elemental sulfur is insoluble in water and crashes out of the solution as a pale yellow precipitate. This is our mysterious "Precipitate X". The balanced equation for this step is:

$$	ext{KI}_3 + 	ext{H}_2	ext{S} ightarrow 	ext{S} \downarrow + 	ext{KI} + 2	ext{HI}$$

### The Final Calculation: Weighing Precipitate X

The stoichiometry of this second reaction is beautifully simple: $1 	ext{ mole}$ of $	ext{KI}_3$ yields exactly $1 	ext{ mole}$ of sulfur precipitate.

Since we had $0.01 	ext{ moles}$ of $	ext{KI}_3$, we will form exactly $0.01 	ext{ moles}$ of sulfur. To find the final answer, we just need to convert these moles back into grams using the atomic mass of sulfur ($32 	ext{ g/mol}$):

$$	ext{Mass of X} = 0.01 	ext{ mol} 	imes 32 	ext{ g/mol} = \mathbf{0.32 	ext{ g}}$$

And there we have it! By carefully tracking the moles through two sequential redox reactions, we have successfully determined the mass of the final precipitate.

Enter Numerical Value:

Visualized Solution

$Visualizing the Reaction Sequence$

$Moles of Cu(NO_{3})_{2}$

$First Reaction: Formation of KI_{3}$

$Moles of KI_{3} Produced$

$Second Reaction: Formation of Precipitate X$

$Mass of Precipitate X$

$The Way Forward$

The Sigma Insight: Stoichiometric and Volumetric Calculations

Unraveling the Mystery of Precipitate X

The Setup

The First Reaction

The Second Reaction

The Final Calculation

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Comprehension Passage

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Reaction Sequence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Cu(NO3​)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Reaction: Formation of KI3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of KI3​ Produced

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Reaction: Formation of Precipitate X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of Precipitate X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward