Answer:
Approximately [tex]1.7\; {\rm L}[/tex].
Explanation:
Nitrogen [tex]{\rm N_{2}}\, (g)[/tex] reacts with hydrogen [tex]{\rm H_{2}}\, (g)[/tex] at a [tex]1:3[/tex] ratio to produce ammonia [tex]{\rm NH_3}\, (g)[/tex]:
[tex]{\rm N_{2}}\, (g) + 3\; {\rm H_{2}}\, (g) \to 2\; {\rm NH_{3}}\, (g)[/tex].
The ratio between the coefficient of [tex]{\rm N_{2}}\, (g)[/tex] and the coefficient of [tex]{\rm H_{2}}\, (g)[/tex] is:
[tex]\begin{aligned}\frac{n({\rm N_{2}})}{n({\rm H_{2}})} = \frac{1}{3}\end{aligned}[/tex].
Under the ideal gas assumptions, the same ratio would apply to the volume of [tex]{\rm N_{2}}\, (g)[/tex] and [tex]{\rm H_{2}}\, (g)[/tex] in this reaction:
[tex]\begin{aligned}\frac{V({\rm N_{2}})}{V({\rm H_{2}})} = \frac{n({\rm N_{2}})}{n({\rm H_{2}})} = \frac{1}{3}\end{aligned}[/tex].
[tex]\begin{aligned}V({\rm N_{2}})= \frac{1}{3}\, V({\rm H_{2}})\end{aligned}[/tex].
Given that [tex]V({\rm H_{2}}) = 5.0\; {\rm L}[/tex]:
[tex]\begin{aligned}V({\rm N_{2}}) &= \frac{1}{3}\, V({\rm H_{2}}) \\ &= \frac{1}{3}\times 5.0\; {\rm L} \\ &\approx 1.7\; {\rm L}\end{aligned}[/tex].
(Rounded to [tex]2[/tex] significant figures.)
