# 320 g of solid ammonium nitrite, NH_4NO_2, decomposes when heated according to the balanced equation: NH_4NO_2 -> N_2+2H_2O. What total volume of gases at 819 K is emitted by this reaction?

Mar 21, 2016

Each mole of $N {H}_{4} N {O}_{2}$ produces $3$ $m o l$ of gases total (${N}_{2}$ and ${H}_{2} O$), and we have $5$ $m o l$. So $3 \times 5 = 15$ $m o l$ of gas at STP is $336$ $L$. Converting to $819$ $K$ gives a volume of $1008$ $L$.

#### Explanation:

The molar mass of $N {H}_{4} N {O}_{2}$ is $64$ $g m o {l}^{-} 1$ (two $N$ at $14$, four $H$ at $1$, two $O$ at $16$).

Find the number of moles of $N {H}_{4} N {O}_{2}$:

$n = \frac{m}{M} = \frac{320}{64} = 5$ $m o l$

From the balanced equation, each mole of $N {H}_{4} N {O}_{2}$ yields 1 mole of ${N}_{2}$ and 2 moles of ${H}_{2} O$ (which is definitely a gas at $819$ $K$), for a total of 3 moles of gas.

For ideal gases (which we can treat these real gases as for our purposes), 1 mole of any gas at STP ($273$ $K$ and $1$ $a t m$) occupies $22.4$ $L$.

That means we have $3 \times 5 = 15$ $m o l$ of product gases, and they occupy a volume of $336$ $L$ at STP.

We need to change the temperature from $273$ $K$ to $819$ $K$, and the pressure stays the same at $1$ $a t m$, so:

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Rearranging:

${V}_{2} = {V}_{1} {T}_{2} / {T}_{1} = 336 \times \frac{819}{273} = 1008$ $L$