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 = 52u, Ba = 137u)

We have to find, for every one gram of the original solid or gas taken as reactant, what mass of nitrogen gas $$N_2$$ is liberated. In other words, for each reaction we shall evaluate

$$\frac{\text{mass of }N_2\text{ obtained}}{\text{mass of reactant consumed}}$$

The reaction giving the largest value of this ratio will be the one that produces the “maximum quantity” of $$N_2$$ per gram of reactant. Let us examine the four reactions one by one.

Reaction (A)  $$\mathrm{Ba(N_3)_2(s)\;\longrightarrow\;Ba(s)+3\,N_2(g)}$$

Molar mass of the reactant $$\mathrm{Ba(N_3)_2$$:

We use the given atomic masses: $$\mathrm{Ba}=137\;u,\;N=14\;u.$$

$$\begin{aligned} M_r[\mathrm{Ba(N_3)_2}]&=&137 \;(\text{from Ba})+2\times(3\times14)\;(\text{from 2 azide ions})\\ &=&137+2\times42\\ &=&137+84\\ &=&221\;\text{g mol}^{-1}. \end{aligned}$$

From the balanced equation, $$1$$ mole of the azide produces $$3$$ moles of $$N_2$$. Hence the mass of $$N_2$$ formed by 1 mole of reactant is

$$m_{N_2}=3\times28=84\;\text{g}.$$ So,

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{84}{221}=0.380\;(\text{approximately}).$$

Reaction (B)  $$\mathrm{(NH_4)_2Cr_2O_7(s)\;\longrightarrow\;N_2(g)+4H_2O(g)+Cr_2O_3(s)}$$

Molar mass of $$\mathrm{(NH_4)_2Cr_2O_7}$$:

Using $$\mathrm{N}=14,\;H=1,\;Cr=52,\;O=16$$,

$$\begin{aligned} M_r&=&2\times14 + 8\times1 + 2\times52 + 7\times16\\ &=&28 + 8 + 104 + 112\\ &=&252\;\text{g mol}^{-1}. \end{aligned}$$

One mole of the dichromate gives exactly one mole of $$N_2$$, i.e.

$$m_{N_2}=1\times28=28\;\text{g}.$$

Therefore

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{28}{252}=0.111.$$

Reaction (C)  $$2NH_3(g)\;\longrightarrow\;N_2(g)+3H_2(g)$$

Molar mass of $$NH_3$$ is

$$M_r[NH_3]=14+3\times1=17\;\text{g mol}^{-1}.$$

The balanced equation shows that $$2$$ moles of $$NH_3$$ (mass $$2\times17=34\;\text{g}$$) produce $$1$$ mole of $$N_2$$ (mass $$28\;\text{g}$$). Hence, for the stoichiometric amount,

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{28}{34}=0.8235\;(\text{approximately}).$$

Because this ratio is for 34 g of reactant, for 1 g the value is the same decimal 0.8235. This is clearly larger than the corresponding ratios found so far.

Reaction (D)  $$2NH_4NO_3(s)\;\longrightarrow\;2N_2(g)+4H_2O(g)+O_2(g)$$

Molar mass of $$NH_4NO_3$$:

$$\begin{aligned} M_r[NH_4NO_3]&=&(2\times14)+(4\times1)+(3\times16)\\ &=&28+4+48\\ &=&80\;\text{g mol}^{-1}. \end{aligned}$$

The reaction consumes $$2$$ moles of nitrate (mass $$2\times80=160\;\text{g}$$) and yields $$2$$ moles of $$N_2$$ (mass $$2\times28=56\;\text{g}$$). Therefore,

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{56}{160}=0.35.$$

Now let us collect the four ratios:

$$ \begin{aligned} \text{Reaction A}:&\;0.380\\ \text{Reaction B}:&\;0.111\\ \text{Reaction C}:&\;\mathbf{0.824}\;(\text{highest})\\ \text{Reaction D}:&\;0.35 \end{aligned} $$

Clearly, Reaction (C) produces the largest mass of $$N_2$$ per gram of reactant.

Hence, the correct answer is Option C.