A sample of ammonium nitrate when heated yields 8.96 L of steam (measured at STP).
`NH_(4)NO_(3) to N_(2)O+2H_(2)O`
What volume of dinitrogen oxide is produced at the same time as 8.96 L of steam?

Explanation

To solve the problem, we need to determine the volume of dinitrogen oxide (N₂O) produced when 8.96 L of steam (H₂O) is generated from the decomposition of ammonium nitrate (NH₄NO₃).

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The decomposition of ammonium nitrate can be represented as:
\[ NH_4NO_3 \rightarrow N_2O + 2H_2O \]
From the equation, we can see that 1 mole of NH₄NO₃ produces 1 mole of N₂O and 2 moles of H₂O.

2. **Calculate the moles of steam produced:**
We are given that 8.96 L of steam is produced at STP (Standard Temperature and Pressure). At STP, 1 mole of any ideal gas occupies 22.4 L. Therefore, we can calculate the number of moles of steam (H₂O) produced:
\[ \text{Moles of } H_2O = \frac{\text{Volume of } H_2O}{\text{Molar Volume at STP}} = \frac{8.96 \, \text{L}}{22.4 \, \text{L/mol}} = 0.4 \, \text{moles} \]

3. **Determine the moles of ammonium nitrate used:**
According to the stoichiometry of the reaction, 2 moles of steam are produced for every 1 mole of NH₄NO₃. Therefore, the moles of NH₄NO₃ required to produce 0.4 moles of steam can be calculated as follows:
\[ \text{Moles of } NH_4NO_3 = \frac{0.4 \, \text{moles of } H_2O}{2} = 0.2 \, \text{moles} \]

4. **Calculate the moles of dinitrogen oxide produced:**
From the balanced equation, we see that 1 mole of NH₄NO₃ produces 1 mole of N₂O. Therefore, the moles of N₂O produced will be equal to the moles of NH₄NO₃ used:
\[ \text{Moles of } N_2O = 0.2 \, \text{moles} \]

5. **Calculate the volume of dinitrogen oxide produced:**
Now, we can find the volume of N₂O produced at STP using the same molar volume:
\[ \text{Volume of } N_2O = \text{Moles of } N_2O \times \text{Molar Volume at STP} = 0.2 \, \text{moles} \times 22.4 \, \text{L/mol} = 4.48 \, \text{L} \]

### Final Answer:
The volume of dinitrogen oxide (N₂O) produced is **4.48 L**.