What total volume , in litre at `627^(@)`C and 0.821 atm, could be
formed by the decomposition of 16 gm of `NH_(3)NO_(3)` ?Reaction : `2NH_(4)NO_(3) to 2N_(2) + O_(2) + 4H_(2)O_((g))` .

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The decomposition of ammonium nitrate (NH₄NO₃) is given by the equation:
\[ 2 \text{NH}_4\text{NO}_3 \rightarrow 2 \text{N}_2 + \text{O}_2 + 4 \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of NH₄NO₃.
The molar mass of NH₄NO₃ can be calculated as follows:
- Nitrogen (N): 14 g/mol (2 Nitrogens = 28 g)
- Hydrogen (H): 1 g/mol (4 Hydrogens = 4 g)
- Oxygen (O): 16 g/mol (3 Oxygens = 48 g)

Total molar mass of NH₄NO₃:
\[ 14 + 4 + 16 + 16 = 80 \text{ g/mol} \]

### Step 3: Calculate the number of moles of NH₄NO₃ in 16 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of NH}_4\text{NO}_3 = \frac{16 \text{ g}}{80 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 4: Determine the number of moles of gas produced.
From the balanced equation, 2 moles of NH₄NO₃ produce 7 moles of gas (2 N₂ + 1 O₂ + 4 H₂O). Therefore, the number of moles of gas produced from 0.2 moles of NH₄NO₃ is:
\[ \text{Moles of gas} = 0.2 \text{ moles NH}_4\text{NO}_3 \times \frac{7 \text{ moles gas}}{2 \text{ moles NH}_4\text{NO}_3} = 0.7 \text{ moles gas} \]

### Step 5: Use the Ideal Gas Law to find the volume.
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (0.821 atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles (0.7 moles)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (627°C + 273 = 900 K)

Rearranging for volume \( V \):
\[ V = \frac{nRT}{P} \]

Substituting the values:
\[ V = \frac{0.7 \text{ moles} \times 0.0821 \text{ L·atm/(K·mol)} \times 900 \text{ K}}{0.821 \text{ atm}} \]

### Step 6: Calculate the volume.
Calculating the volume:
\[ V = \frac{0.7 \times 0.0821 \times 900}{0.821} \]
\[ V \approx \frac{42.927}{0.821} \approx 52.2 \text{ liters} \]

### Final Answer:
The total volume of gas produced is approximately **52.2 liters**.

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