Decomposition of ammonium chloride is an endothermic reaction. The equilibrium may be represented as:
`NH_(4)Cl(s) hArr NH_(3)(g)+HCl(g)`
A `6.250 g` sample of `NH_(4)Cl` os placed in an evaculated `4.0 L` container at `27^(@)C`. After equilibrium the total pressure inside the container is `0.820` bar and some solid remains in the container. Answer the followings
The amount of solid `NH_(4)Cl` left behind in the container at equilibrium is

Options

A. `2.856`
B. `28.56`
C. `0.2856`
D. `1.320`

Explanation

To solve the problem, we need to determine the amount of solid NH₄Cl left in the container at equilibrium after the decomposition reaction has occurred. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the Equilibrium Reaction
The decomposition of ammonium chloride can be represented as:
\[ \text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCl}(g) \]

### Step 2: Calculate the Total Moles of Gases at Equilibrium
We can use the ideal gas law to find the total number of moles of gases (NH₃ and HCl) at equilibrium. The formula is:
\[ n = \frac{PV}{RT} \]
Where:
- \( P = 0.820 \, \text{bar} \) (total pressure)
- \( V = 4.0 \, \text{L} \) (volume)
- \( R = 0.0821 \, \text{L·bar/(K·mol)} \) (ideal gas constant)
- \( T = 27°C = 300 \, \text{K} \) (temperature in Kelvin)

### Step 3: Substitute Values into the Ideal Gas Law
Substituting the values into the equation:
\[ n = \frac{0.820 \, \text{bar} \times 4.0 \, \text{L}}{0.0821 \, \text{L·bar/(K·mol)} \times 300 \, \text{K}} \]

### Step 4: Calculate the Number of Moles
Calculating this gives:
\[ n = \frac{3.28}{24.63} \approx 0.133 \, \text{moles} \]
This is the total number of moles of NH₃ and HCl at equilibrium.

### Step 5: Define the Change in Moles
Let \( \alpha \) be the number of moles of NH₄Cl that dissociate. At equilibrium:
- Moles of NH₃ = \( \alpha \)
- Moles of HCl = \( \alpha \)
- Remaining moles of NH₄Cl = \( n - \alpha \)

Since both gases are produced in a 1:1 ratio:
\[ \alpha + \alpha = 0.133 \]
Thus:
\[ 2\alpha = 0.133 \]
\[ \alpha = 0.0665 \, \text{moles} \]

### Step 6: Calculate Initial Moles of NH₄Cl
Next, we need to calculate the initial moles of NH₄Cl present:
- Given mass of NH₄Cl = 6.250 g
- Molar mass of NH₄Cl = 53.5 g/mol

Calculating the moles:
\[ n = \frac{6.250 \, \text{g}}{53.5 \, \text{g/mol}} \approx 0.116 \, \text{moles} \]

### Step 7: Calculate Remaining Moles of NH₄Cl
Now, we can find the remaining moles of NH₄Cl at equilibrium:
\[ \text{Remaining moles of NH}_4\text{Cl} = n - \alpha = 0.116 - 0.0665 = 0.0495 \, \text{moles} \]

### Step 8: Convert Moles to Mass
Finally, convert the remaining moles of NH₄Cl back to grams:
\[ \text{Mass} = \text{moles} \times \text{molar mass} = 0.0495 \, \text{moles} \times 53.5 \, \text{g/mol} \approx 2.65 \, \text{g} \]

### Conclusion
The amount of solid NH₄Cl left behind in the container at equilibrium is approximately **2.65 g**.

---