Sure! Let's approach this problem step-by-step using stoichiometry based on the balanced chemical equation:
[tex]$
N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)
$[/tex]
1. Identify the volume of NH₃ given:
- We are given that the volume of ammonia ([tex]\( NH_3 \)[/tex]) produced is 446 liters.
2. Understand the stoichiometric relationship:
- According to the balanced chemical equation, 1 volume of [tex]\( N_2 \)[/tex] reacts with 3 volumes of [tex]\( H_2 \)[/tex] to produce 2 volumes of [tex]\( NH_3 \)[/tex].
- Therefore, the stoichiometric ratio of [tex]\( H_2 \)[/tex] to [tex]\( NH_3 \)[/tex] is [tex]\( \frac{3 \, \text{vol H}_2}{2 \, \text{vol NH}_3} \)[/tex].
3. Set up the proportion using the stoichiometric ratio:
- From the stoichiometric ratio, we know that for every 2 volumes of [tex]\( NH_3 \)[/tex] produced, 3 volumes of [tex]\( H_2 \)[/tex] are consumed.
- Hence, the volume of [tex]\( H_2 \)[/tex] needed can be calculated as:
[tex]$
\text{Volume of } H_2 = \left( \frac{3 \text{ vol H}_2}{2 \text{ vol } NH_3} \right) \times \text{Volume of } NH_3
$[/tex]
4. Plug in the volume of [tex]\( NH_3 \)[/tex] into the equation:
- Since the volume of [tex]\( NH_3 \)[/tex] is 446 liters, we substitute it in the equation:
[tex]$
\text{Volume of } H_2 = \left( \frac{3}{2} \right) \times 446 \text{ L}
$[/tex]
5. Perform the multiplication:
- Calculate:
[tex]$
\text{Volume of } H_2 = 1.5 \times 446 \text{ L} = 669 \text{ L}
$[/tex]
Therefore, the volume of [tex]\( H_2 \)[/tex] needed to generate 446 liters of [tex]\( NH_3 \)[/tex] at STP is 669 liters.
