# Question #a7871

Aug 10, 2017

Approx......$1 \cdot L$........

#### Explanation:

We gots.......

$N {H}_{4} N {O}_{2} \left(s\right) \stackrel{\Delta}{\rightarrow} {N}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

And thus there is 1:1 molar equivalence between ammonium nitrite, and dinitrogen gas.....

$\text{Moles of ammonium nitrite} = \frac{2.25 \cdot g}{64.06 \cdot g \cdot m o {l}^{-} 1} = 0.0351 \cdot m o l$.

And this we evolve $0.0351 \cdot m o l$ dinitrogen gas......

Now $V = \frac{n R T}{P} = \frac{0.0351 \cdot m o l \cdot 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 299.2 \cdot K}{\frac{745 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}}$

$\cong 0.9 \cdot L$

The key to answering this problem is to know that $1 \cdot a t m$ pressure will support a column of mercury that is $760 \cdot m m$ high. And thus we can use the height of a mercury column as a very good VISUAL representation of pressure.