A solid soluble in water on being heated gives brown gas and when dil. HCI is added to its aqueous solution, a white ppt is obtained. The solid is
A solid soluble in water on being heated gives brown gas and when dil. HCI is added to its aqueous solution, a white ppt is obtained. The solid is
A
`Ca(NO_(3))_2`
B
`Zn(NO_(3))`
C
`Pb(NO_(3))_2`
D
`Ba(NO_(3))_2`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem step by step, we need to analyze the information provided in the question:
### Step 1: Identify the properties of the solid
The solid is described as being soluble in water and producing a brown gas upon heating.
**Hint:** Think about common solids that are soluble in water and can release gases when heated.
### Step 2: Determine the brown gas produced
When the solid is heated, it produces a brown gas. A common brown gas that can be produced from the decomposition of certain compounds is nitrogen dioxide (NO2).
**Hint:** Recall the gases produced from the thermal decomposition of nitrates.
### Step 3: Consider the chemical reaction
Lead(II) nitrate (Pb(NO3)2) is known to decompose upon heating to produce lead(II) oxide (PbO), nitrogen dioxide (NO2), and oxygen (O2):
\[ \text{2 Pb(NO}_3\text{)}_2 \rightarrow \text{2 PbO} + \text{4 NO}_2 + \text{O}_2 \]
This reaction confirms that heating lead(II) nitrate produces a brown gas (NO2).
**Hint:** Look for a nitrate compound that fits the description provided in the question.
### Step 4: Analyze the reaction with dilute HCl
The question states that when dilute hydrochloric acid (HCl) is added to the aqueous solution of the solid, a white precipitate is formed. Lead(II) chloride (PbCl2) is a common white precipitate formed when lead(II) ions react with chloride ions.
**Hint:** Consider the reactions that occur when lead(II) ions are combined with chloride ions.
### Step 5: Write the reaction with dilute HCl
When lead(II) nitrate is dissolved in water and then treated with dilute HCl, the following reaction occurs:
\[ \text{Pb(NO}_3\text{)}_2 + 2 \text{HCl} \rightarrow \text{PbCl}_2 \downarrow + 2 \text{HNO}_3 \]
This reaction produces a white precipitate of lead(II) chloride (PbCl2) and nitric acid (HNO3).
**Hint:** Identify the products formed when lead(II) nitrate reacts with hydrochloric acid.
### Conclusion
Based on the analysis, the solid that meets all the criteria described in the question is **lead(II) nitrate (Pb(NO3)2)**.
### Final Answer
The solid is **Pb(NO3)2** (lead(II) nitrate).
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