Question

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. - $Cr=52\, u, Ba=137 \,u$)

• A. $\left( NH _{4}\right)_{2} Cr _{2} O _{7}( s ) \rightarrow N _{2}( g )+4 H _{2} O ( g )+ Cr _{2} O _{3}( s )$
• B. $2 NH _{4} NO _{3}( s ) \rightarrow 2 N _{2}( g )+4 H _{2} O ( g )+ O _{2}( g )$
• C. $Ba \left( N _{3}\right)_{2}( s ) \rightarrow Ba ( s )+3 N _{2}( g )$
• D. $2 NH _{3}( g ) \rightarrow N _{2}( g )+3 H _{2}( g )$

Answer 1

Answer: (D)

Option (1) : Molar mass of $( NH _{4} \text { ) } Cr _{2} O _{7}=252\, g \,mol ^{-1}$
1 mol of $\left( NH _{4}\right)_{2} Cr _{2} O _{7} $ will give 1 mol of $N _{2}$
$\frac{1}{252} $ mol of $\left( NH _{4}\right)_{2} Cr _{2} O _{7}$ will give $ \frac{1}{252} \times 1=0.039$ mol of $ N _{2}$
Option (2): Molar mass of $ NH _{4} NO _{3}=80 \,g \,mol ^{-1}$
1 mol of $NH _{4} NO _{3}$ will give 1 mol of $N _{2}$
$\frac{1}{80} $ mol of $NH _{4} NO _{3}$ will give $ \frac{1}{80} \times 1=0.0125 $ mol of $ N _{2}$
Option (3): Molar mass of $Ba \left( N _{3}\right)_{2}=221 \,g \,mol ^{-1}$
1 mol of $Ba \left( N _{3}\right)_{2} $ will give 3 mol of $N_2$
$\frac{1}{221}$ mol of $ Ba \left( N _{3}\right)_{2}$ will give $\frac{1}{221} \times 3=0.014$ mol of $ N _{2}$
Option (4): Molar mass of $NH _{3}=17 \,g \,mol ^{-3}$
2 mol of $NH _{3} $ will give 1 mol of $N _{2}$
$\frac{1}{17} $ mol of $NH _{3}$ will give $\frac{1}{2 \times 17}=0.0297 $ mol of $ N _{2}$