We need to compare the amount of $N_2$ produced per gram of reactant for each reaction.

That means for each option, compute:

$\text{moles of }N_2\text{ produced per gram of reactant} = \frac{\text{moles of }N_2\text{ formed per stoichiometric amount}}{\text{molar mass of stoichiometric amount of reactant}}$

Since volume or mass of $N_2$ is directly proportional to moles, the option with maximum moles of $N_2$ per gram will be correct.

1. Option A

$(NH_4)_2Cr_2O_7 \to N_2 + 4H_2O + Cr_2O_3$

Step 1: Molar mass of $(NH_4)_2Cr_2O_7$

• $N$: $2 \times 14 = 28$
• $H$: $8 \times 1 = 8$
• $Cr$: $2 \times 52 = 104$
• $O$: $7 \times 16 = 112$

So,

$M = 28 + 8 + 104 + 112 = 252\,\text{g mol}^{-1}$

Step 2: $N_2$ formed

From $1$ mole reactant, $1$ mole $N_2$ is formed.

So moles of $N_2$ per gram:

$\frac{1}{252}$

2. Option B

$2NH_4NO_3 \to 2N_2 + 4H_2O + O_2$

Step 1: Molar mass of $NH_4NO_3$

• $N$: $2 \times 14 = 28$
• $H$: $4 \times 1 = 4$
• $O$: $3 \times 16 = 48$

So,

$M = 28 + 4 + 48 = 80\,\text{g mol}^{-1}$

For $2$ moles, mass $= 160$ g.

Step 2: $N_2$ formed

From $2$ moles reactant, $2$ moles $N_2$ are formed.

So moles of $N_2$ per gram:

$\frac{2}{160} = \frac{1}{80}$

3. Option C

$Ba(N_3)_2 \to Ba + 3N_2$

Step 1: Molar mass of $Ba(N_3)_2$

$Ba(N_3)_2$ contains $BaN_6$.

• $Ba$: $137$
• $N$: $6 \times 14 = 84$

So,

$M = 137 + 84 = 221\,\text{g mol}^{-1}$

Step 2: $N_2$ formed

From $1$ mole reactant, $3$ moles $N_2$ are formed.

So moles of $N_2$ per gram:

$\frac{3}{221}$

4. Option D

$2NH_3 \to N_2 + 3H_2$

Step 1: Molar mass of $NH_3$

$M(NH_3)=14+3=17\,\text{g mol}^{-1}$

For $2$ moles, mass $=34$ g.

Step 2: $N_2$ formed

From $2$ moles reactant, $1$ mole $N_2$ is formed.

So moles of $N_2$ per gram:

$\frac{1}{34}$

5. Compare all options

Now compare:

• A: $\dfrac{1}{252} \approx 0.00397$
• B: $\dfrac{1}{80} = 0.0125$
• C: $\dfrac{3}{221} \approx 0.01357$
• D: $\dfrac{1}{34} \approx 0.02941$

Clearly,

$\frac{1}{34} > \frac{3}{221} > \frac{1}{80} > \frac{1}{252}$

So the maximum quantity of $N_2$ per gram of reactant is produced in Option D.

Final Answer

Option D is correct.