`A overset(Delta)rarrB` (oxide) `+ CO_(2)`
`B + H_(2)O rarr C`
`C+ CO_(2) rarrA` (milky)
`C+ NH_(4)CI overset(Delta)rarr D` (gas)
`D + H_(2)O + CO_(2) rarr E`
`E + NaCI rarr F`
`F overset(Delta)rarr Na_(2)CO_(3)+CO_(2)+H_(2)O`
D,E and F are:

Options

A. `NH_(3), NH_(4)CI, NH_(4)HCO_(3)`
B. `NH_(3), NH_(4)HCO_(3), NaHCO_(3)`
C. `NH_(4)HCO_(3), Na_(2)CO_(3), NaHCO_(3)`
D. None

Answer

To solve the problem, we need to identify the substances D, E, and F based on the given series of reactions. Let's break down the reactions step by step.

### Step 1: Analyze the last reaction
The last reaction given is:
\[ F \overset{\Delta}{\rightarrow} \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]

From this reaction, we can deduce that F is likely a precursor to sodium carbonate (Na2CO3).

### Step 2: Determine F
To find F, we can reverse the reaction:
\[ \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow F \]

Balancing this reaction, we find that:
\[ F = 2 \text{NaHCO}_3 \]
Thus, **F is sodium bicarbonate (NaHCO3)**.

### Step 3: Analyze the reaction involving E
The next reaction is:
\[ E + \text{NaCl} \rightarrow F \]

Since we established that F is NaHCO3, we can write:
\[ E + \text{NaCl} \rightarrow 2 \text{NaHCO}_3 \]

To find E, we can rearrange:
\[ E = 2 \text{NaHCO}_3 - \text{NaCl} \]

This reaction suggests that E is ammonium bicarbonate (NH4HCO3). Thus, **E is ammonium bicarbonate (NH4HCO3)**.

### Step 4: Analyze the reaction involving D
Next, we look at:
\[ D + \text{H}_2\text{O} + \text{CO}_2 \rightarrow E \]

Substituting E:
\[ D + \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{NH}_4\text{HCO}_3 \]

Rearranging gives:
\[ D = \text{NH}_3 \]

Thus, **D is ammonia (NH3)**.

### Summary of Findings
- **D** is ammonia (NH3).
- **E** is ammonium bicarbonate (NH4HCO3).
- **F** is sodium bicarbonate (NaHCO3).

### Final Answer
D, E, and F are:
- D = NH3 (ammonia)
- E = NH4HCO3 (ammonium bicarbonate)
- F = NaHCO3 (sodium bicarbonate)