To determine how many water molecules are produced when 33.0 g of dinitrogen oxide (N₂O) is produced, we first need to use stoichiometry based on the balanced chemical equation provided:
\[ \text{NH}_4\text{NO}_3 (s) \rightarrow \text{N}_2\text{O} (g) + 2 \text{H}_2\text{O} (l) \]
Step 1: Calculate the moles of N₂O
First, we calculate the molar mass of dinitrogen oxide (N₂O):
- Molar mass of N₂O = (2 × Molar mass of N) + (1 × Molar mass of O)
- Molar mass of N₂O = (2 × 14.0 g/mol) + (1 × 16.0 g/mol) = 28.0 g/mol + 16.0 g/mol = 44.0 g/mol
Now, we find the number of moles of N₂O produced:
\[ \text{Moles of N₂O} = \frac{33.0 , \text{g}}{44.0 , \text{g/mol}} = 0.75 , \text{mol} \]
Step 2: Use the stoichiometry of the reaction
According to the balanced reaction, 1 mole of N₂O produces 2 moles of water (H₂O). Thus,
\[ \text{Moles of H₂O} = 0.75 , \text{mol N₂O} \times 2 , \frac{\text{mol H₂O}}{\text{mol N₂O}} = 1.5 , \text{mol H₂O} \]
Step 3: Calculate the number of water molecules
To find the number of molecules, we use Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol):
\[ \text{Number of H₂O molecules} = 1.5 , \text{mol H₂O} \times 6.022 \times 10^{23} , \text{molecules/mol} \]
Calculating this gives:
\[ \text{Number of H₂O molecules} = 1.5 \times 6.022 \times 10^{23} \approx 9.033 \times 10^{23} , \text{molecules} \]
Final Answer
Thus, when 33.0 g of N₂O is produced, approximately \(9.03 \times 10^{23}\) water molecules are produced.
1 answer