Ammonium nitrate decomposes when heated to produce dinitrogen monoxide and water, as follows:

\[ \text{NH}_4\text{NO}_3(s) \rightarrow \text{N}_2\text{O}(g) + \text{H}_2\text{O}(l) \]

If 12.2 g of $\text{NH}_4\text{NO}_3$ reacts, what volume of $\text{N}_2\text{O}$, measured at 98.5 kPa and 14$^\circ\text{C}$, is produced?

To calculate the volume of dinitrogen monoxide produced from 12.2 g of ammonium nitrate at 98.5 kPa and 14ºC, we use the ideal gas law and stoichiometry, which gives us a volume of 3.80 L.

Explanation:

A student asked how to calculate the volume of dinitrogen monoxide (N₂O) produced when 12.2 g of ammonium nitrate (NH4NO3) decomposes at a given temperature and pressure. To find this, we will use the ideal gas law, Stoichiometry, and the molar mass of ammonium nitrate.

First, calculate the number of moles of NH4NO3 using its molar mass (80.043 g/mol).

12.2 g NH4NO3 × (1 mol NH4NO3 / 80.043 g NH4NO3) = 0.1524 moles NH4NO3

Using the balanced equation NH4NO3 (s) → N₂O (g) + 2H₂O (l), we can see that 1 mole of NH4NO3 produces 1 mole of N₂O. Thus,
0.1524 moles NH4NO3 → 0.1524 moles N₂O

Now, using the ideal gas law PV=nRT, where
P = 98.5 kPa (convert to atm),
V = ?,
n = 0.1524 moles,
R = 0.0821 L·atm/mol·K,
T = 14ºC (convert to Kelvin).
First, convert P to atm: 98.5 kPa × (1 atm / 101.325 kPa) = 0.9718 atm.
Then, convert T to Kelvin: 14ºC + 273.15 = 287.15 K.
V = (nRT)/P = (0.1524 mol × 0.0821 L·atm/mol·K × 287.15 K) / 0.9718 atm = 3.80 L of N₂O gas.