What volume of N2 is required to convert 5.0L of hydrogen gas to ammonia? assume that all gases are at the same temperature and pressure and that the reaction is complete

Question

asked Dec 26, 2023 213k views

1 Answer

Samir Adel · Answer 1 · 2024-01-01T18:20:54+0000

Answer:

Approximately
$1.7\; {\rm L}$ .

Step-by-step explanation:

Nitrogen
${\rm N_(2)}\, (g)$ reacts with hydrogen
${\rm H_(2)}\, (g)$ at a
1:3 ratio to produce ammonia
${\rm NH_3}\, (g)$ :

${\rm N_(2)}\, (g) + 3\; {\rm H_(2)}\, (g) \to 2\; {\rm NH_(3)}\, (g)$ .

The ratio between the coefficient of
${\rm N_(2)}\, (g)$ and the coefficient of
${\rm H_(2)}\, (g)$ is:

$\begin{aligned}\frac{n({\rm N_(2)})}{n({\rm H_(2)})} = (1)/(3)\end{aligned}$ .

Under the ideal gas assumptions, the same ratio would apply to the volume of
${\rm N_(2)}\, (g)$ and
${\rm H_(2)}\, (g)$ in this reaction:

$\begin{aligned}\frac{V({\rm N_(2)})}{V({\rm H_(2)})} = \frac{n({\rm N_(2)})}{n({\rm H_(2)})} = (1)/(3)\end{aligned}$ .

$\begin{aligned}V({\rm N_(2)})= (1)/(3)\, V({\rm H_(2)})\end{aligned}$ .

Given that
$V({\rm H_(2)}) = 5.0\; {\rm L}$ :

$\begin{aligned}V({\rm N_(2)}) &= (1)/(3)\, V({\rm H_(2)}) \\ &= (1)/(3)* 5.0\; {\rm L} \\ &\approx 1.7\; {\rm L}\end{aligned}$ .

(Rounded to
significant figures.)