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 mass of ammonium nitrate should be heated to produce 8.96 L of steam?
(Relative molecular mass of ammonium nitrate is 80.)

Explanation

To solve the problem, we need to determine the mass of ammonium nitrate (NH₄NO₃) required to produce 8.96 L of steam when heated. The reaction is given as:

\[ \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O} \]

### Step-by-Step Solution:

1. **Calculate the moles of steam produced**:
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Therefore, we can calculate the moles of steam produced from the given volume of steam (8.96 L).

\[ \text{Moles of steam} = \frac{\text{Volume of steam}}{22.4 \, \text{L/mol}} = \frac{8.96 \, \text{L}}{22.4 \, \text{L/mol}} = 0.4 \, \text{mol} \]

2. **Use stoichiometry to find moles of ammonium nitrate**:
From the balanced equation, we see that 1 mole of ammonium nitrate produces 2 moles of steam. Therefore, the moles of ammonium nitrate required can be calculated as follows:

\[ \text{Moles of NH}_4\text{NO}_3 = \frac{1}{2} \times \text{Moles of steam} = \frac{1}{2} \times 0.4 \, \text{mol} = 0.2 \, \text{mol} \]

3. **Calculate the mass of ammonium nitrate**:
We know the molar mass of ammonium nitrate (NH₄NO₃) is given as 80 g/mol. We can use the number of moles to find the mass:

\[ \text{Mass of NH}_4\text{NO}_3 = \text{Moles} \times \text{Molar mass} = 0.2 \, \text{mol} \times 80 \, \text{g/mol} = 16 \, \text{g} \]

### Final Answer:
The mass of ammonium nitrate that should be heated to produce 8.96 L of steam is **16 grams**.

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