In the presence of nitric acid, \(UO\) \(^{2+}\) undergoes a redox process. It is converted to \(\mathrm{UO}_{2}^{2+}\) and nitric oxide (NO) gas is produced according to the following unbalanced equation: $$\begin{aligned}\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2 *}(a q) & \longrightarrow \\\\\mathrm{NO}(g)+& \mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ If \(2.55 \times 10^{2} \mathrm{mL} \mathrm{NO}(g)\) is isolated at \(29^{\circ} \mathrm{C}\) and \(1.5 \mathrm{atm}\), what amount (moles) of \(UO\) \(^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)

We will balance the given redox equation step by step: 1. Determine the oxidation states of each element: \(H^+\) : +1 \(NO_3^-\) : N = +5, O = -2 \(UO^{2+}\) : U = +6, O = -2 \(NO\) : N = +2, O = -2 \(UO_2^{2+}\) : U = +4, O = -2 \(H_2O\): H = +1, O = -2 2. Identify the elements undergoing oxidation and reduction: - U is reduced (oxidation state change: +6 to +4) - N in \(NO_3^-\) is oxidized (oxidation state change: +5 to +2) 3. Balance the half-reactions: - Reduction half-reaction: \(UO^{2+} \rightarrow UO_2^{2+}\) - To balance O, add 1 H_2O: \(UO^{2+} \rightarrow UO_2^{2+} + H_2O\) - Oxidation half-reaction: \(NO_3^- \rightarrow NO\) - To balance O, add 2 H^+: \(NO_3^- + 2H^+ \rightarrow NO + H_2O\) 4. Balance the charge by adding electrons and balance the number of electrons transferred in both half-reactions: - Reduction half-reaction: \(UO^{2+} + 2e^{-} \rightarrow UO_2^{2+} + H_2O\) - Oxidation half-reaction: \(NO_3^- + 2H^+ \rightarrow NO + H_2O + 3e^{-}\) - To balance electrons, multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2: - \(3(UO^{2+} + 2e^{-} \rightarrow UO_2^{2+} + H_2O)\) - \(2(NO_3^- + 2H^+ \rightarrow NO + H_2O + 3e^{-})\) 5. Combine the balanced half-reactions and simplify: - \(6H^+ + 2NO_3^- + 3UO^{2+} \rightarrow 3UO_2^{2+} + 4H_2O + 2NO\) The balanced redox equation is: \(6H^+ + 2NO_3^- + 3UO^{2+} \rightarrow 3UO_2^{2+} + 4H_2O + 2NO\)

Given the volume, temperature, and pressure of NO gas, use the Ideal Gas Law, \(PV = nRT\), where P is pressure, V is volume, n is moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. Convert temperature to Kelvin and volume to Liters: - \(T = 29°C + 273.15 = 302.15K\) - \(V = 2.55 × 10^2 mL = 0.255 L\) Solve for n: \(n = \frac{PV}{RT} = \frac{(1.5 atm)(0.255 L)}{(0.0821 L·atm/mol·K)(302.15 K)} = 0.0126 mol\) There are 0.0126 moles of NO gas produced.

Using the balanced redox equation and stoichiometry, we can determine the moles of \(UO^{2+}\) used: - From the balanced equation: \(3UO^{2+} \rightarrow 2NO\) - Calculate the moles of \(UO^{2+}\) used based on the moles of NO produced: \(\frac{3 \ moles\ UO^{2+}}{2 \ moles\ NO} × 0.0126 \ moles\ NO = 0.0189 \ moles\ UO^{2+}\) There were 0.0189 moles of \(UO^{2+}\) used in the reaction.