Certainly! Let's solve this problem step by step.
We start with the balanced chemical equation for the reaction:
[tex]\[
N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)
\][/tex]
This equation tells us that 1 mole of nitrogen gas ([tex]\( N_2 \)[/tex]) reacts with 3 moles of hydrogen gas ([tex]\( H_2 \)[/tex]) to produce 2 moles of ammonia ([tex]\( NH_3 \)[/tex]).
From the problem, we know we need to generate 12 liters of ammonia ([tex]\( NH_3 \)[/tex]) at standard temperature and pressure (STP).
According to Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules, we can directly use the stoichiometric ratios given by the balanced equation because the relationship between volumes of gases in a chemical reaction at STP is the same as the relationship between moles.
From the balanced equation, we see the stoichiometric ratio:
[tex]\[
\text{Volumes of } H_2 : \text{Volumes of } NH_3 = 3 : 2
\][/tex]
This tells us that 3 volumes of hydrogen produce 2 volumes of ammonia. Let's denote the volume of hydrogen needed as [tex]\( V_{H_2} \)[/tex].
Given that we need to produce 12 liters of ammonia, we can set up the proportion based on the stoichiometric ratio:
[tex]\[
\frac{V_{H_2}}{12 \, \text{liters of } NH_3} = \frac{3}{2}
\][/tex]
Solving for [tex]\( V_{H_2} \)[/tex], we get:
[tex]\[
V_{H_2} = 12 \, \text{liters} \times \frac{3}{2}
\][/tex]
[tex]\[
V_{H_2} = 18 \, \text{liters}
\][/tex]
Therefore, the volume of hydrogen gas ([tex]\( H_2 \)[/tex]) needed to generate 12 liters of ammonia ([tex]\( NH_3 \)[/tex]) at STP is:
[tex]\[
\boxed{18 \, \text{liters}}
\][/tex]
