Laughing gas is sometimes used as an anesthetic in dentistry. It is produced when ammonium nitrate decomposes to form dinitrogen

Question

oxide and water. Use the reaction below to answer the following questions.
Type of atom Molar Mass (g/mol)
N 14.0
O 16.0
H 1.01
NH4NO3 (s) → N2O (g) + 2 H2 O (l)
If 33.0 g of N2O is produced, how many water molecules will be produced?

Answer 1

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:

$${\text{NH}}_4{\text{NO}}_3(s)\to {\text{N}}_2\text{O}(g)+2{\text{H}}_2\text{O}(l)$$

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.