To determine how many moles of oxygen (O₂) are produced from the decomposition of 36.7 grams of dinitrogen pentoxide (N₂O₅), we need to follow these steps:
1. Calculate the molar mass of dinitrogen pentoxide (N₂O₅):
- Dinitrogen pentoxide comprises 2 nitrogen (N) atoms and 5 oxygen (O) atoms.
- The atomic mass of nitrogen (N) is approximately 14 g/mol.
- The atomic mass of oxygen (O) is approximately 16 g/mol.
- Molar mass of N₂O₅ = 2(14) + 5(16) = 28 + 80 = 108 g/mol.
2. Determine the number of moles of dinitrogen pentoxide present in 36.7 grams:
- Moles of N₂O₅ = [tex]\(\frac{\text{mass}}{\text{molar mass}} = \frac{36.7 \text{ g}}{108 \text{ g/mol}}\)[/tex].
- Moles of N₂O₅ ≈ 0.3398 mol (accurate to 4 decimal places).
3. Use the stoichiometric relationship from the balanced chemical equation to calculate the moles of oxygen produced:
- The balanced equation is: [tex]\(2 \text{N}_2\text{O}_5 \rightarrow 4 \text{NO}_2 + \text{O}_2\)[/tex].
- According to the equation, 2 moles of N₂O₅ produce 1 mole of O₂.
- Therefore, the number of moles of O₂ produced = [tex]\(\frac{1}{2} \times \)[/tex] number of moles of N₂O₅.
4. Calculate the moles of oxygen produced:
- Moles of O₂ = [tex]\(\frac{1}{2} \times 0.3398 \text{ mol}\)[/tex].
- Moles of O₂ ≈ 0.1699 mol (accurate to 4 decimal places).
Hence, the number of moles of oxygen produced from the decomposition of 36.7 grams of dinitrogen pentoxide is approximately 0.170 mol.
The correct answer is:
B. 0.170 mol
