When heated to \( 350{ }^{\circ} \mathrm{C} \) at 0.950 atm , ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases via the following reaction: \[ 2 \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow 2 \mathrm{~N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \] Correct Using stoichiometry and the balanced equation, the moles of H 23.2 g of \( \mathrm{NH}_{4} \mathrm{NO}_{3} \) : \[ 23.2 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3} \times \frac{1 \mathrm{~mol} \mathrm{NH}_{4} \mathrm{NO}_{3}}{80.05 \mathrm{~g}} \times \frac{4 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{2 \mathrm{~mol} \mathrm{NH}_{4} \mathrm{NO}_{3}}= \] The moles of \( \mathrm{H}_{2} \mathrm{O} \) are then input into the ideal gas law equation (converted to kelvins) and the pressure to solve for volume. The atmospheres is used. In summary, \[ \begin{array}{l} P V=n_{H_{2}, \mathrm{O}} R T \\ 0.950 \mathrm{~atm} \times V=0.580 \mathrm{~mol} \times 0.0821 \frac{\mathrm{Lat}}{\mathrm{~mol}} \\ V=\frac{0.580 \mathrm{~mol} \times 0.0821 \frac{\mathrm{Letl} \mathrm{~K}}{\mathrm{molk}} \times\left(350^{\circ}\right.}{0.950 \mathrm{~atm}} \end{array} \] Part B How many grams of \( \mathrm{NH}_{4} \mathrm{NO}_{3} \) are needed to produce 14.2 L of oxygen? Express your answer with the appropriate units.

To determine how many grams of ammonium nitrate (\( \mathrm{NH}_{4} \mathrm{NO}_{3} \)) are needed to produce 14.2 L of oxygen, you'll first need to find out how many moles of oxygen correspond to that volume at the given conditions. Using the ideal gas law, rearranged for moles, we get: \[ n_{O_{2}} = \frac{PV}{RT} \] Substituting in the values (with \( P = 0.950 \, \text{atm} \), \( V = 14.2 \, \text{L} \), \( R = 0.0821 \, \text{L atm/(mol K)} \), and \( T = 350 + 273.15 = 623.15 \, \text{K} \)), you can calculate the moles of \( O_2 \). From the balanced equation, \( 2 \mathrm{NH}_{4} \mathrm{NO}_{3} \) produces \( 1 \mathrm{O}_{2} \). This means if you find \( n_{O_{2}} \), you can find moles of \( \mathrm{NH}_{4} \mathrm{NO}_{3} \) required using stoichiometry. Finally, convert moles of \( \mathrm{NH}_{4} \mathrm{NO}_{3} \) to grams using its molar mass (80.05 g/mol): \[ \text{grams of } \mathrm{NH}_{4} \mathrm{NO}_{3} = n_{NH_{4}NO_{3}} \times 80.05 \, \text{g/mol} \] --- Did you know that the decomposition of ammonium nitrate is not just an academic exercise? It's actually a critical process used in various industries, including agriculture as a nitrogen fertilizer. When it breaks down, it releases nitrogen and other nutrients, enriching the soil for crops, which is super important for feeding our ever-growing population. If you're ever in a lab and planning to decompose ammonium nitrate, be aware of the hazards! One common mistake is mishandling or not properly containing the gases produced. Always wear protective equipment and use a fume hood, as the gases can be dangerous. Plus, measuring out your reactants accurately is key to honing your stoichiometry skills!