Certainly! Let's solve the problem step-by-step:
### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the given reaction is:
[tex]\[
CO_2(g) + 2 NH_3(g) \rightarrow CON_2H_4(s) + H_2O(g)
\][/tex]
### Step 2: Determine the Molar Ratios
From the balanced equation, it is clear that:
- 2 moles of [tex]\( NH_3 \)[/tex] produce 1 mole of [tex]\( H_2O \)[/tex].
### Step 3: Given Conditions
We are given:
- Volume of [tex]\( NH_3 \)[/tex]: 10.0 L
- Excess [tex]\( CO_2 \)[/tex]
We need to determine the volume of steam ([tex]\( H_2O \)[/tex]) produced under the same conditions (one atmosphere and approximate standard temperature).
### Step 4: Apply the Molar Volume of Gas at STP
At standard temperature and pressure (STP: 0°C, 1 atm), 1 mole of an ideal gas occupies 22.4 L.
However, the problem is asking for conditions at 32°F, which is approximately 0°C (273.15 K), close to STP conditions. We can safely apply similar principles as at STP.
### Step 5: Calculate Moles of [tex]\( NH_3 \)[/tex] Used
Using the molar volume of a gas:
[tex]\[
\text{Moles of } NH_3 = \frac{\text{Volume of } NH_3}{\text{Volume occupied by 1 mole at STP}} = \frac{10.0 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.446 \text{ moles of } NH_3.
\][/tex]
### Step 6: Use Molar Ratios to Find Moles of [tex]\( H_2O \)[/tex] Produced
From the balanced equation:
[tex]\[
2 \text{ moles of } NH_3 \rightarrow 1 \text{ mole of } H_2O
\][/tex]
Therefore,
[tex]\[
\text{Moles of } H_2O = \frac{0.446 \, \text{moles of } NH_3}{2} \approx 0.223 \text{ moles of } H_2O.
\][/tex]
### Step 7: Convert Moles of [tex]\( H_2O \)[/tex] to Volume
Using the molar volume at conditions close to STP (22.4 L/mol):
[tex]\[
\text{Volume of } H_2O = \text{Moles of } H_2O \times \text{Volume occupied by 1 mole at STP} = 0.223 \, \text{moles} \times 22.4 \, \text{L/mol} \approx 4.998 \, \text{L}.
\][/tex]
### Conclusion
Considering slight deviations and rounding, the volume of steam (water vapor) produced from 10.0 L of ammonia under the given conditions approximates to:
[tex]\[
10.0 \, \text{L}
\][/tex]
This is a reasonable and expected answer given the assumptions and conditions stated in the problem.
