The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make $1000.0 \mathrm{~L}\( of a \)0.150 \mathrm{M}$ aqueous solution of nitric acid? Assume all the reactions give \(100 \%\) yield.

To identify redox reactions, we'll look at the oxidation states of each element in reactants and products. If the oxidation state changes, then the reaction is a redox reaction. To determine the oxidation states, remember that hydrogen has oxidation state +1, oxygen has oxidation state -2, and the oxidative state of nitrogen varies. For the first reaction: \[4NH_3(g)+5O_2(g) \longrightarrow 4NO(g)+6H_2O(g)\] Oxidation states (initial and final) are: N: -3 to +2 H: +1 (stays constant) O: 0 to -2 For the second reaction: \[2NO(g)+O_2(g) \longrightarrow 2NO_2(g)\] Oxidation states (initial and final) are: N: +2 to +4 O: 0 to -2 For the third reaction: \[3NO_2(g)+H_2O(l) \longrightarrow 2HNO_3(aq)+NO(g)\] Oxidation states (initial and final) are: N: +4 to +5 O: -2 (stays constant) H: +1 (stays constant) As there are changes in oxidation states for all three reactions, all three are redox reactions.

Based on the oxidation state changes from Step 1, we can determine which elements are undergoing oxidation (increase in oxidation state) and reduction (decrease in oxidation state): For the first reaction: Oxidation: Nitrogen (N) - going from -3 to +2 Reduction: Oxygen (O) - going from 0 to -2 For the second reaction: Oxidation: Nitrogen (N) - going from +2 to +4 Reduction: Oxygen (O) - going from 0 to -2 For the third reaction: Oxidation: Nitrogen (N) - going from +4 to +5 Reduction: None, since no elements decrease in oxidation state.

To find the amount of ammonia needed to produce 1000.0 L of a 0.150 M aqueous solution of nitric acid, first, we need to find the number of moles of nitric acid in this solution: Moles of \(HNO_3\) = Molarity × Volume Moles of \(HNO_3\) = \(0.150\, mol/L \times 1000.0\, L = 150.0\, mol\) Now, look at the balanced equation of the first and third reaction: Reaction 1: \(4NH_3 + 5O_2 \longrightarrow 4NO + 6H_2O\) Reaction 3: \(3NO_2 + H_2O \longrightarrow 2HNO_3 + NO\) Since all reactions have 100% yield, we can calculate the moles of ammonia needed for 150.0 moles of \(HNO_3\). From Reaction 3, we can see that 2 moles of \(HNO_3\) are produced from 1 mole of NO. From Reaction 1, we can see that 4 moles of NO are produced from 4 moles of \(NH_3\). Thus, we can find the moles of ammonia (\(NH_3\)) needed: Moles of \(NH_3 = \frac{moles \, of \, HNO_3}{2} \times 4 = 150.0\, mol \times 2 = 300.0\, mol\) Finally, convert moles of ammonia to grams using the molar mass of ammonia (17.03 g/mol): Mass of \(NH_3 =\) moles of \(NH_3 \times\) molar mass of \(NH_3\) Mass of \(NH_3 = 300.0\, mol \times 17.03\, g/mol = 5109\, g\) Thus, 5109 grams of ammonia are needed to make 1000.0 L of a 0.150 M aqueous solution of nitric acid assuming 100% yield.