Ammonium nitrite, \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), decomposes according to the chemical equation below. $$\mathrm{NH}_{4} \mathrm{NO}_{2}(\mathrm{s}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What is the total volume of products obtained when \(128 \mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{2}\) decomposes at \(819^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa} ?\)

First calculate the number of moles of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) using the formula \(\mathrm{Moles} = \frac{\mathrm{Mass}}{\mathrm{Molar mass}}\). From the periodic table, the molar mass of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) is approximately \(32 + 14 + 16 * 2 = 78 \, \mathrm{g/mol}\). Thus, the number of moles of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) is \(\frac{128 \, \mathrm{g}}{78 \, \mathrm{g/mol}} \approx 1.64 \, \mathrm{moles}\).

From the stoichiometric coefficients in the balanced chemical reaction, one mole of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) produces one mole of \(\mathrm{N}_{2}\) gas and 2 moles of \(\mathrm{H}_{2} \mathrm{O}\) gas. Therefore, the total moles of gas produced is \(1.64 \, \mathrm{moles} * (1 + 2) = 4.92 \, \mathrm{moles}\).

The ideal gas law is \(\mathrm{PV} = \mathrm{nRT}\), where \(\mathrm{P}\) is the pressure in \(\mathrm{kPa}\), \(\mathrm{V}\) is the volume in \(\mathrm{l}\), \(\mathrm{n}\) is the moles of gas, \(\mathrm{R}\) is the ideal gas constant = \(8.31 \, \mathrm{kPa \cdot L \cdot K^{-1} \cdot mol^{-1}}\), and \(\mathrm{T}\) is the Kelvin temperature. First convert the temperature from Celsius to Kelvin by adding 273, giving \(\mathrm{T} = 819 + 273 = 1092 \, \mathrm{K}\). Substituting the given values into the ideal gas law, the volume \(\mathrm{V}\) is obtained by \(\mathrm{V} = \frac{\mathrm{nRT}}{\mathrm{P}} = \frac{4.92 \, \mathrm{moles} * 8.31 \, \mathrm{kPa \cdot L \cdot K^{-1} \cdot mol^{-1}} * 1092 \, \mathrm{K}}{101 \, \mathrm{kPa}} \approx 406.9 \, \mathrm{L}\).