To solve the problem, we need to determine the volume of ammonia (NH₃) required to produce 1575 g of nitric acid (HNO₃) through a three-step process with given percent yields. Let's break it down step by step.
### Step 1: Calculate the number of moles of HNO₃ produced
First, we need to find the molar mass of HNO₃:
- H: 1 g/mol
- N: 14 g/mol
- O: 16 g/mol × 3 = 48 g/mol
So, the molar mass of HNO₃ = 1 + 14 + 48 = 63 g/mol.
Now, we can calculate the number of moles of HNO₃ produced from 1575 g:
\[
\text{Moles of HNO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{1575 \, \text{g}}{63 \, \text{g/mol}} \approx 25 \, \text{mol}
\]
### Step 2: Work backwards through the reactions to find moles of NH₃ needed
**Step 3: Third Reaction (HNO₃ production)**
The third reaction is:
\[
3 \, \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2 \, \text{HNO}_3(aq) + \text{NO}(g)
\]
From this reaction, we see that 2 moles of HNO₃ are produced from 3 moles of NO₂. Therefore, to produce 25 moles of HNO₃, we need:
\[
\text{Moles of NO}_2 = \frac{3}{2} \times 25 = 37.5 \, \text{mol}
\]
Considering the yield of the third step is 80%, the actual moles of NO₂ required will be:
\[
\text{Required moles of NO}_2 = \frac{37.5}{0.8} = 46.875 \, \text{mol}
\]
**Step 4: Second Reaction (NO production)**
The second reaction is:
\[
2 \, \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \, \text{NO}_3(g)
\]
From this reaction, 2 moles of NO produce 2 moles of NO₃. Therefore, to produce 46.875 moles of NO₃, we need:
\[
\text{Moles of NO} = 46.875 \, \text{mol}
\]
Considering the yield of the second step is 60%, the actual moles of NO required will be:
\[
\text{Required moles of NO} = \frac{46.875}{0.6} = 78.125 \, \text{mol}
\]
**Step 5: First Reaction (NH₃ production)**
The first reaction is:
\[
4 \, \text{NH}_3(g) + 5 \, \text{O}_2(g) \rightarrow 4 \, \text{NO}(g) + 6 \, \text{H}_2\text{O}(g)
\]
From this reaction, 4 moles of NH₃ produce 4 moles of NO. Therefore, to produce 78.125 moles of NO, we need:
\[
\text{Moles of NH}_3 = 78.125 \, \text{mol}
\]
Since the yield of the first step is 50%, the actual moles of NH₃ required will be:
\[
\text{Required moles of NH}_3 = \frac{78.125}{0.5} = 156.25 \, \text{mol}
\]
### Step 6: Calculate the volume of NH₃ at STP
At standard temperature and pressure (STP: 0°C and 1 atm), 1 mole of gas occupies 22.4 liters. Therefore, the volume of NH₃ required is:
\[
\text{Volume of NH}_3 = 156.25 \, \text{mol} \times 22.4 \, \text{L/mol} = 3500 \, \text{L}
\]
### Final Answer
The volume of NH₃ required to produce 1575 g of HNO₃ is **3500 liters**.
