Answer:
To solve this problem, we need to follow these steps:
1. **Balance the chemical equation:** The reaction given is already balanced:
[tex]\[ \text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O} \][/tex]
2. **Molecular weight calculation:** Calculate the molar mass of ammonium nitrite [tex](\(\text{NH}_4\text{NO}_2\))[/tex].
[tex]\[ \text{Molar mass of } \text{NH}_4\text{NO}_2 = 14.01 (\text{N}) + 4 \times 1.01 (\text{H}) + 14.01 (\text{N}) + 2 \times 16.00 (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of } \text{NH}_4\text{NO}_2 = 14.01 + 4.04 + 14.01 + 32.00 = 64.06 \text{ g/mol} \][/tex]
3. **Convert grams to moles:** Using the molar mass, convert the mass of [tex]\(\text{NH}_4\text{NO}_2\)[/tex] used to moles:
[tex]\[ \text{Moles of } \text{NH}_4\text{NO}_2 = \frac{493.83 \text{ g}}{64.06 \text{ g/mol}} \approx 7.71 \text{ moles} \][/tex]
4. **Mole ratio calculation:** According to the balanced equation, 1 mole of [tex]\(\text{NH}_4\text{NO}_2\)[/tex] produces 1 mole of [tex]\(\text{N}_2\)[/tex]. Therefore, 7.71 moles of [tex]\(\text{NH}_4\text{NO}_2\)[/tex] will produce 7.71 moles of [tex]\(\text{N}_2\)[/tex].
5. **Volume of nitrogen gas at STP:** At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. So, the volume of nitrogen gas produced can be calculated as:
[tex]\[ \text{Volume of } \text{N}_2 = 7.71 \text{ moles} \times 22.4 \text{ L/mole} \][/tex]
[tex]\[ \text{Volume of } \text{N}_2 = 172.70 \text{ liters} \][/tex]
Therefore, if 493.83 grams of [tex]\(\text{NH}_4\text{NO}_2\)[/tex] are reacted, approximately **172.70 liters** of nitrogen gas will be produced at STP.
Explanation:
