Question

What is the total volume (in $L$) at ${627}^{\circ }C$ and $1$ atm that could be formed by the decomposition of $16g$ of $N{H}_{4}N{O}_3$?
Reaction : $2N{H}_{4}N{O}_3\to 2N_2+O_2+4H_{2}O\left(g\right)$

• A. $57.47l$
• B. $114.94ml$
• C. $41.78l$
• D. $24.63l$

Answer 1

The correct option is A $57.47l$

The molar mass of ammonium nitrate is $80g/mol$.
$2$ moles of ammonium nitrate on decomposition gives $7$ moles of product.
Hence, $16g$ of ammonium nitrate on decomposition will give $\frac{16\times 7}{80\times 2}=0.7$ moles of product.
This corresponds to $0.7\times 22.4=15.68L$ at STP.
At ${627}^{\circ }C$, the total volume is equal to $V_2=\frac{T_2}{T_1}\times V_1=\frac{900}{273}\times 15.68=51.723L$ .