For per gram of reactant, the maximum quantity of `N_(2)` gas is produced in which of the following thermal decomposition reactions?
(Given: Atomic wt. Cr = 52, u, Ba = 137 u)

Options

A. `2NH_(4)NO_(3(s)) rarr 2N_(2(g)) + 4H_(2)O_((g)) + O_(2(g))`
B. `Ba(N_(3))_(2(s)) rarr Ba_((s)) + 3N_(2(g))`
C. `(NH_(4))_(2)Cr_(2)O_(7(s)) rarr N_(2(g)) + 4H_(2)O_((g)) + Cr_(2)O_(3(s))`
D. `2NH_(3(g)) rarr N_(2(g)) + 3H_(2(g))`

Explanation

To determine which thermal decomposition reaction produces the maximum quantity of nitrogen gas (N₂) per gram of reactant, we will analyze each option step by step.

### Step 1: Analyze Ammonium Nitrate (NH₄NO₃)
1. **Calculate Molar Mass**:
- Molar mass of NH₄NO₃ = 14 (N) + 1*4 (H) + 14 (N) + 16*3 (O) = 80 g/mol.
2. **Reaction**:
- 2 NH₄NO₃ → 2 N₂ + 4 H₂O.
- 1 mole of NH₄NO₃ produces 1 mole of N₂.
3. **Calculate N₂ produced per gram**:
- For 1 g of NH₄NO₃:
\[ \text{Moles of NH₄NO₃} = \frac{1 \text{ g}}{80 \text{ g/mol}} = 0.0125 \text{ moles} \]
- Therefore, N₂ produced = 0.0125 moles.

### Step 2: Analyze Barium Azide (Ba(N₃)₂)
1. **Calculate Molar Mass**:
- Molar mass of Ba(N₃)₂ = 137 (Ba) + 2*(14*3) (N) = 221 g/mol.
2. **Reaction**:
- Ba(N₃)₂ → Ba + 3 N₂.
- 1 mole of Ba(N₃)₂ produces 3 moles of N₂.
3. **Calculate N₂ produced per gram**:
- For 1 g of Ba(N₃)₂:
\[ \text{Moles of Ba(N₃)₂} = \frac{1 \text{ g}}{221 \text{ g/mol}} = 0.00452 \text{ moles} \]
- Therefore, N₂ produced = 3 * 0.00452 = 0.01356 moles.

### Step 3: Analyze Ammonium Dichromate (NH₄)₂Cr₂O₇
1. **Calculate Molar Mass**:
- Molar mass of (NH₄)₂Cr₂O₇ = 2*(14 + 4*1) + 2*52 + 7*16 = 252 g/mol.
2. **Reaction**:
- (NH₄)₂Cr₂O₇ → Cr₂O₃ + N₂ + 4 H₂O.
- 1 mole of (NH₄)₂Cr₂O₇ produces 1 mole of N₂.
3. **Calculate N₂ produced per gram**:
- For 1 g of (NH₄)₂Cr₂O₇:
\[ \text{Moles of (NH₄)₂Cr₂O₇} = \frac{1 \text{ g}}{252 \text{ g/mol}} = 0.00397 \text{ moles} \]
- Therefore, N₂ produced = 0.00397 moles.

### Step 4: Analyze Ammonia (NH₃)
1. **Calculate Molar Mass**:
- Molar mass of NH₃ = 14 (N) + 3*1 (H) = 17 g/mol.
2. **Reaction**:
- 2 NH₃ → N₂ + 3 H₂.
- 2 moles of NH₃ produce 1 mole of N₂.
3. **Calculate N₂ produced per gram**:
- For 1 g of NH₃:
\[ \text{Moles of NH₃} = \frac{1 \text{ g}}{17 \text{ g/mol}} = 0.0588 \text{ moles} \]
- Therefore, N₂ produced = 0.0588 / 2 = 0.0294 moles.

### Conclusion:
After calculating the moles of N₂ produced per gram of each reactant:
- Ammonium Nitrate: 0.0125 moles
- Barium Azide: 0.01356 moles
- Ammonium Dichromate: 0.00397 moles
- Ammonia: 0.0294 moles

The maximum quantity of N₂ gas produced per gram of reactant is from **Ammonia (NH₃)**.

### Final Answer:
**Ammonia (NH₃)** produces the maximum quantity of nitrogen gas per gram of reactant.