Question

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.)

Short answer

The amount of UO^2+ used in the reaction is \(3.07 \times 10^{-3} mol\).

Step-by-step solution

Balance the given redox reaction

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

Use the Ideal Gas Law to find the number of moles of NO produced

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: \(n = 3.07 \times 10^{-3} mol\)

Use the stoichiometric relationship to find the moles of UO^2+ used

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 \(3.07 \times 10^{-3} mol\).

Key concepts

Oxidation States

Understanding oxidation states is crucial in balancing redox reactions. In a redox reaction, the oxidation states of elements change as they either gain or lose electrons. This change helps in identifying which element is oxidized and which is reduced. Oxidation can be thought of as the process of losing electrons, exemplified by an increase in oxidation state. Conversely, reduction involves gaining electrons, denoted by a decrease in oxidation state.

Let's look at the exercise: Assigning oxidation states helps in balancing the reaction mathematically. Here are some basic rules:

• Hydrogen generally has an oxidation state of +1.
• Oxygen usually has an oxidation state of -2.
• For ions, the sum of oxidation states must equal the charge of the ion.

In our example, Uranium in \(\mathrm{UO}^{2+}\) has an oxidation state of +4, changing to +6 in \(\mathrm{UO}_{2}^{2+}\) when reduced. Nitrogen in \(\mathrm{NO}^{3-}\) with an oxidation state of +5 is reduced to +2 in \(\mathrm{NO}\), showing a clear change depicting reduction. By understanding these changes, we can efficiently balance the overall reaction.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry used to relate the pressure, volume, temperature, and moles of a gas. It is encapsulated in the equation \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles, \(R\) is the universal gas constant, and \(T\) represents temperature in Kelvin.

In the given exercise, we apply the Ideal Gas Law to find the number of moles of nitric oxide (NO) gas produced. Before using the formula, remember to:

• Convert volume from milliliters to liters.
• Convert temperature from Celsius to Kelvin.

Plug these values into the formula: \[1.5 \, \text{atm} \times 2.55 \times 10^{-1} \, \text{L} = n \times 0.08206 \, \text{L atm/(K mol)} \times 302.15 \, \text{K}\]Upon solving the equation, you'll find the number of moles of \(\mathrm{NO}\) to be \(3.07 \times 10^{-3} \mathrm{mol}\). This approach makes it possible to quantify gaseous reactants and products in chemical reactions.

Stoichiometry

Stoichiometry involves using a balanced chemical equation to calculate the quantities of reactants and products. It serves as a bridge connecting the numerical amount of substances reacting and the stoichiometric coefficients from their balanced equation.

In our redox reaction, the stoichiometric relationship is derived from the balanced equation: \(2 \mathrm{UO}^{2+} + 2 \mathrm{NO}_{3}^{-} + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{UO}_{2}^{2+} + 2 \mathrm{NO} + 4 \mathrm{H}_{2}\mathrm{O}\). From this, the stoichiometric ratio of \(\mathrm{UO}^{2+}\) to \(\mathrm{NO}\) is 1:1.

The stoichiometry tells us that moles of \(\mathrm{UO}^{2+}\) consumed equal the moles of \(\mathrm{NO}\) produced. Given that the moles of \(\mathrm{NO}\) were calculated to be \(3.07 \times 10^{-3} \) mol using the Ideal Gas Law, we can directly conclude that the same number of moles of \(\mathrm{UO}^{2+}\) reacted. Stoichiometry offers a systematic way to solve complex chemical equation problems efficiently.