To determine how many grams of nitrogen ([tex]\( \text{N}_2 \)[/tex]) are required to produce 100.0 liters of ammonia ([tex]\( \text{NH}_3 \)[/tex]) at standard temperature and pressure (STP), we can follow these steps:
### Step 1: Identify the Molar Volume at STP
At STP, one mole of any ideal gas occupies 22.4 liters. Thus, we can use this value to find the number of moles of [tex]\( \text{NH}_3 \)[/tex].
### Step 2: Determine the Moles of [tex]\( \text{NH}_3 \)[/tex]
Given 100.0 liters of [tex]\( \text{NH}_3 \)[/tex] at STP, we can calculate the moles of [tex]\( \text{NH}_3 \)[/tex]:
[tex]\[ \text{moles of } \text{NH}_3 = \frac{\text{volume of } \text{NH}_3}{\text{molar volume at STP}} = \frac{100.0 \, \text{L}}{22.4 \, \text{L/mol}} \approx 4.464 \, \text{moles} \][/tex]
### Step 3: Use Stoichiometry of the Reaction
The balanced chemical equation for the reaction is:
[tex]\[ \text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g) \][/tex]
From the equation, we see that 1 mole of [tex]\( \text{N}_2 \)[/tex] produces 2 moles of [tex]\( \text{NH}_3 \)[/tex]. Therefore, we need:
[tex]\[ \text{moles of } \text{N}_2 = \frac{\text{moles of } \text{NH}_3}{2} = \frac{4.464}{2} \approx 2.232 \, \text{moles} \][/tex]
### Step 4: Calculate the Mass of [tex]\( \text{N}_2 \)[/tex]
We know the molar mass of [tex]\( \text{N}_2 \)[/tex] is 28.0 grams per mole. Using this, we can calculate the mass of [tex]\( \text{N}_2 \)[/tex] required:
[tex]\[ \text{mass of } \text{N}_2 = \text{moles of } \text{N}_2 \times \text{molar mass of } \text{N}_2 = 2.232 \, \text{moles} \times 28.0 \, \text{g/mol} = 62.5 \, \text{g} \][/tex]
### Conclusion
The number of grams of nitrogen required to produce 100.0 liters of ammonia at STP is:
[tex]\[ 62.5 \, \text{grams} \][/tex]
Thus, the correct answer is:
[tex]\[ 62.5 \, \text{g} \][/tex]
