Question

NH4NO3 → N2O + 2H2O
Question: How many grams of ammonium nitrate are needed to produce 12.0 grams of
N2O (dinitrogen monoxide)?

Answer 1

To determine how many grams of ammonium nitrate (NH4NO3) are required to produce 12.0 grams of dinitrogen monoxide (N2O), we need to follow several steps involving stoichiometry.

1. **Write the balanced chemical equation**:
\[
2 \, \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\, \text{H}_2\text{O}
\]

2. **Determine the molar mass of N2O**:
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol

\[
\text{Molar mass of } \text{N}_2\text{O} = (2 \times 14.01) + (1 \times 16.00) = 28.02 + 16.00 = 44.02 \, \text{g/mol}
\]

3. **Convert grams of N2O to moles**:
\[
\text{Moles of N}_2\text{O} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{12.0 \, \text{g}}{44.02 \, \text{g/mol}} \approx 0.272 \, \text{mol}
\]

4. **Use the stoichiometry from the balanced equation**:
From the balanced equation, 1 mole of N2O is produced from 2 moles of NH4NO3. Therefore:

\[
\text{Moles of NH}_4\text{NO}_3 = 2 \times \text{Moles of N}_2\text{O} = 2 \times 0.272 \, \text{mol} \approx 0.544 \, \text{mol}
\]

5. **Determine the molar mass of NH4NO3**:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol

\[
\text{Molar mass of } \text{NH}_4\text{NO}_3 = (2 \times 14.01) + (4 \times 1.01) + (3 \times 16.00)
\]
\[
= 28.02 + 4.04 + 48.00 = 80.06 \, \text{g/mol}
\]

6. **Convert moles of NH4NO3 to grams**:
\[
\text{Mass of NH}_4\text{NO}_3 = \text{Moles} \times \text{Molar mass} = 0.544 \, \text{mol} \times 80.06 \, \text{g/mol} \approx 43.55 \, \text{g}
\]

Thus, approximately **43.55 grams of ammonium nitrate are needed to produce 12.0 grams of N2O**.