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: \(\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2+}(a q) \longrightarrow\) \(\mathrm{NO}(g)+\mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) 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

Expert verified
The amount of \(\mathrm{UO}^{2+}\) used in the reaction is 0.0204 moles.

Step by step solution

01

Balance the given redox reaction equation

To balance the redox reaction, first identify the oxidation and reduction half-reactions. The oxidation state of Uranium changes from +4 to +6 and the oxidation state of Nitrogen changes from +5 to +2. The half-reactions are: Oxidation: \(\mathrm{UO}^{2+} \longrightarrow \mathrm{UO}_2^{2+} + 2 e^-\) Reduction: \( \mathrm{NO}_3^- + 3 e^- \longrightarrow \mathrm{NO} + \mathrm{H_2O} \) Now, balance the charge by multiplying the half-reactions by appropriate factors and add them together: \(x3\): \(\mathrm{UO}^{2+} \longrightarrow \mathrm{UO}_2^{2+} + 2 e^-\) \(x2\): \( \mathrm{NO}_3^- + 3 e^- \longrightarrow \mathrm{NO} + \mathrm{H_2O} \) Sum: \(3\mathrm{UO}^{2+} + 2\mathrm{NO}_3^- \longrightarrow 3\mathrm{UO}_2^{2+} + 2\mathrm{NO} + 2\mathrm{H_2O} \) Now add \(\mathrm{H}^+\) ions to balance the hydrogen: \(3\mathrm{UO}^{2+} + 2\mathrm{NO}_3^- + 4 \mathrm{H}^+ \longrightarrow 3\mathrm{UO}_2^{2+} + 2\mathrm{NO} + 2 \mathrm{H_2O}\) The balanced equation is: \(3\mathrm{UO}^{2+} + 2\mathrm{NO}_3^- + 4 \mathrm{H}^+ \longrightarrow 3\mathrm{UO}_2^{2+} + 2\mathrm{NO} + 2 \mathrm{H_2O}\)
02

Calculate the moles of NO gas produced using the ideal gas law

We have the volume, temperature and pressure of the isolated NO gas. The ideal gas law relates these parameters: \(PV = nRT\) Here, P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, which equals \(8.314 \frac{J}{mol\cdot K}\), and T is temperature in Kelvins. First, convert temperature from Celsius to Kelvin: \(T = 29 + 273.15 = 302.15 \thinspace K\) Insert the given values and solve for n (moles of NO): \(1.5 \thinspace atm \times 2.55 \times 10^2 \thinspace mL = n \times 8.314 \frac{J}{mol\cdot K} \times 302.15 \thinspace K\) Before using the ideal gas law, we need to convert units: pressure to \(atm\) and volume to L. \(Pressure = 1.5 \thinspace atm\) (already in atm) \(Volume = 2.55 \times 10^2 \thinspace mL = 0.255 \thinspace L\) Now, we can plug the values into the ideal gas equation: \(1.5 \thinspace atm \times 0.255 \thinspace L = n \times 0.0821 \frac{L\cdot atm}{mol\cdot K} \times 302.15 \thinspace K\) Solve for n: \(n_{NO} = 0.0136 \thinspace mol.\)
03

Determine the amount (moles) of UO^{2+} used in the reaction

Now, using the stoichiometry of the balanced equation, determine the amount of moles of \(\mathrm{UO}^{2+}\) used in the reaction. From the balanced equation, we have: \(3\mathrm{UO}^{2+} + 2\mathrm{NO}_3^- + 4 \mathrm{H}^+ \longrightarrow 3\mathrm{UO}_2^{2+} + 2\mathrm{NO} + 2 \mathrm{H_2O}\) The stoichiometry shows that 2 moles of NO are produced for every 3 moles of \(\mathrm{UO}^{2+}\) that react. Set up the proportion and solve for moles of \(\mathrm{UO}^{2+}\) used: \(\frac{3\thinspace moles \thinspace of \thinspace UO^{2+}}{2\thinspace moles \thinspace of \thinspace NO} = \frac{x\thinspace moles \thinspace of \thinspace UO^{2+}}{0.0136\thinspace moles \thinspace of \thinspace NO}\) Solve for x: \(x = 0.0204 \thinspace moles \thinspace of \thinspace UO^{2+}\) The amount of \(\mathrm{UO}^{2+}\) used in the reaction is 0.0204 moles.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Oxidation States in Redox Reactions

Redox reactions are fundamental types of chemical reactions where oxidation and reduction processes occur simultaneously. To analyze redox reactions, it is essential to understand oxidation states, as they help indicate the degree of oxidation or reduction of an atom within a compound. In a redox reaction, the substance that loses electrons is oxidized (increase in oxidation state), while the substance that gains electrons is reduced (decrease in oxidation state).


Oxidation states can be determined by following certain rules based on electronegativity, elemental states, and known charges on ions. For instance, in our exercise, the oxidation state of uranium changes from +4 to +6, which means it undergoes oxidation by losing electrons. Conversely, nitrogen goes from +5 to +2, indicating it accepts electrons and is reduced. The importance of correctly identifying oxidation states cannot be overstated, as it is the first step in balancing a redox equation.

The Role of Stoichiometry in Determining Reactant Quantities

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles, which allows chemists to calculate the amounts of substances consumed and produced. When you have a balanced chemical equation, stoichiometry becomes a powerful tool to predict how much of each reactant is needed or how much of a product will be formed.


In the provided exercise, stoichiometry is used to connect the moles of nitric oxide gas produced to the original moles of \(\mathrm{UO}^{2+}\) reacted. Given that the balanced equation shows a 3:2 ratio between \(\mathrm{UO}^{2+}\) and NO, stoichiometry allows us to establish a direct proportional relationship. By knowing the amount of one substance, we can calculate the amount of another. This makes stoichiometry a key concept for solving many practical problems in chemistry, not just in the classroom but also in industrial applications like pharmaceuticals and materials science.

Applying the Ideal Gas Law to Quantify Gases in Reactions

The ideal gas law is a crucial equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas through the formula \(PV = nRT\). It is based on the ideal gas assumption, which treats gases as a collection of particles that have no volume and do not interact, allowing for straightforward calculations. While real gases may deviate from this ideal behavior, the ideal gas law is incredibly useful for estimating the behavior of gases under many conditions.


When applying the ideal gas law, as seen in this exercise, it's important to ensure all units are consistent, typically using liters (L) for volume, atmospheres (atm) for pressure, Kelvin (K) for temperature, and moles (mol) for the amount of substance. The gas constant \(R\) has different values depending on the units used; for the ideal gas law using atm, L, and K, \(R\) equals 0.0821 \((L\cdot atm)/(mol\cdot K)\). By knowing any three of the variables, we can solve for the fourth. This forms the basis for solving a variety of problems, including finding the amount of a gas produced or consumed in a chemical reaction, as demonstrated in the exercise.

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Most popular questions from this chapter

Which of the following statements is(are) true? For the false statements, correct them. a. At constant temperature, the lighter the gas molecules, the faster the average velocity of the gas molecules. b. At constant temperature, the heavier the gas molecules, the larger the average kinetic energy of the gas molecules. c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving relatively quickly. d. As temperature increases, the effect of interparticle interactions on gas behavior is increased. e. At constant \(V\) and \(T\), as gas molecules are added into a container, the number of collisions per unit area increases resulting in a higher pressure. f. The kinetic molecular theory predicts that pressure is inversely proportional to temperature at constant volume and moles of gas.

The rate of effusion of a particular gas was measured and found to be \(24.0 \mathrm{~mL} / \mathrm{min}\). Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is \(47.8 \mathrm{~mL} / \mathrm{min}\). What is the molar mass of the unknown gas?

A mixture of chromium and zinc weighing \(0.362 \mathrm{~g}\) was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, \(225 \mathrm{~mL}\) dry of hydrogen gas was collected at \(27^{\circ} \mathrm{C}\) and 750 . torr. Determine the mass percent of \(\mathrm{Zn}\) in the metal sample. [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium(III) chloride and hydrogen gas.]

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -mL nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another \(100.0\) -mL nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

Consider two separate gas containers at the following conditions: $$ \begin{array}{|ll|} \text { Container A } & \text { Container B } \\ \text { Contents: } \mathrm{SO}_{2}(g) & \text { Contents: unknown gas } \\ \text { Pressure }=P_{\mathrm{A}} & \text { Pressure }=P_{\mathrm{B}} \\ \text { Moles of gas }=1.0 \mathrm{~mol} & \text { Moles of gas }=2.0 \mathrm{~mol} \\ \text { Volume }=1.0 \mathrm{~L} & \text { Volume }=2.0 \mathrm{~L} \\ \text { Temperature }=7^{\circ} \mathrm{C} & \text { Temperature }=287^{\circ} \mathrm{C} \\ \hline \end{array} $$ How is the pressure in container \(\mathrm{B}\) related to the pressure in container \(\mathrm{A}\) ?

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