In the presence of nitric acid, \(\mathrm{UO}^{2+}\) 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: \(\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2+}(a q)\) \(\mathrm{NO}(g)+\mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) 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 \(\mathrm{UO}^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)

Let's balance the unbalanced equation by the oxidation states method. First, assign oxidation states for all elements involved in the reaction: H^+ has an oxidation state of +1 NO3^-: N has an oxidation state of +5, and O has an oxidation state of -2 UO^2+: U has an oxidation state of +4, and O has an oxidation state of -2 NO: N has an oxidation state of +2, and O has an oxidation state of -2 UO2^2+: U has an oxidation state of +6, and O has an oxidation state of -2 H2O: H has an oxidation state of +1, and O has an oxidation state of -2 The balanced half-reactions will be: 2 UO^2+ + 3 H2O ⟶ 2 UO2^2+ + 6 H^+ + 2e^- (Reduction half-reaction) 2 NO3^- + 3 H^+ + 2e^- ⟶ 2 NO + 3 H2O (Oxidation half-reaction) Now, we can add the two half-reactions to obtain the full balanced equation: 2 UO^2+ + 2 NO3^- + 6 H^+ ⟶ 2 UO2^2+ + 2 NO + 4 H2O

With the volume, temperature, and pressure of the nitric oxide given, we can use the Ideal Gas Law to find the number of moles of NO produced: \(PV = nRT\) Where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the universal gas constant (0.08206 L atm/(K mol)), and T is the temperature in Kelvin. First, we need to convert the volume from mL to L and the temperature from Celsius to Kelvin: Volume in L: \(2.55 \times 10^2 mL = 2.55 \times 10^{-1} L\) Temperature in K: \(29^{\circ}C + 273.15 = 302.15 K\) Now, plug in the values and solve for n: \(1.5 atm * 2.55 \times 10^{-1} L = n * 0.08206 L atm/(K mol) * 302.15 K\) Solving for n, we get: \(n = 3.07 \times 10^{-3} mol\)

In the balanced reaction, we can see that the stoichiometric ratio of UO^2+ to NO is 2:2, which simplifies to 1:1. This means that the moles of UO^2+ used in the reaction is equal to the moles of NO produced: Moles of UO^2+ used = Moles of NO produced Moles of UO^2+ used = 3.07 × 10^-3 mol So, the amount of UO^2+ used in the reaction is \(3.07 \times 10^{-3} mol\).