Potassium superoxide, KO 2 , is often used in oxygen masks (such as those used by firefighters) because KO 2 reacts with CO 2 to release molecular oxygen. Experiments indicate that 2 mol of KO 2 (s) react with each mole of CO 2 (g). (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?

Hey everyone, we're asked to give the oxidation number for each element. Starting off with our react inside. Let's look at our platinum for sulfate. So we know that our platinum has a charge of plus four and that's because we require to sulfates in order to balance out that plus four. So platinum has an oxidation state of plus four And we know that oxygen has an oxidation state of -2. And if we solve for X and say sulfate is X, we can simply do X plus four times negative two since we have four oxygen's And we can equal this to negative two since the overall charge of sulfate is -2, solving for X, we get x minus eight equals negative two and X will be plus six. So silver's oxidation state is going to be plus six. Moving on to rubidium iodine, we know that rubidium has an oxidation state of plus one. Since it is a group won a medal, iodine has an oxidation state of -1 since it's a group seven a medal. Looking at our product side, we can see that platinum has now become Plus two in its oxidation state since we have two. I'd eines balancing out our platinum and speaking of iodine, we can see that iodine ended up with an oxidation state of zero since it is now in its standard state. Looking at our sulfur and oxygen's, They still remained plus six and -2 respectively, and rubidium also remained A Plus one Oxidation State. So I hope that made sense. And let us know if you have any questions

Ch.4 - Reactions in Aqueous Solution

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Brown 14th Edition

Ch.4 - Reactions in Aqueous Solution

Problem 115 b

Chapter 4, Problem 115 b

Potassium superoxide, KO₂, is often used in oxygen masks (such as those used by firefighters) because KO₂ reacts with CO₂ to release molecular oxygen. Experiments indicate that 2 mol of KO₂(s) react with each mole of CO₂(g). (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?

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Identify the chemical reaction: 4 KO_2 + 2 CO_2 \rightarrow 2 K_2CO_3 + 3 O_2.

Determine the oxidation number of each element in the reactants: K in KO_2 is +1, O in KO_2 is -1/2, C in CO_2 is +4, and O in CO_2 is -2.

Determine the oxidation number of each element in the products: K in K_2CO_3 is +1, C in K_2CO_3 is +4, O in K_2CO_3 is -2, and O_2 is 0.

Compare the oxidation numbers of each element in the reactants and products to identify changes: O in KO_2 changes from -1/2 to 0, indicating oxidation, and O in CO_2 changes from -2 to -2, indicating no change.

Conclude which elements are oxidized and reduced: Oxygen in KO_2 is oxidized, and no element is reduced as there is no decrease in oxidation number.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Oxidation States

Oxidation states, or oxidation numbers, are a way to keep track of electrons in chemical reactions. They indicate the degree of oxidation of an atom in a compound, helping to identify which atoms are losing or gaining electrons. In the context of the reaction involving KO2 and CO2, determining the oxidation states of potassium, oxygen, and carbon is essential for identifying the elements that are oxidized and reduced.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two species. In these reactions, one species is oxidized (loses electrons) while another is reduced (gains electrons). Understanding the principles of redox reactions is crucial for analyzing the reaction between KO2 and CO2, as it allows us to determine which elements undergo oxidation and reduction.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the conservation of mass. It involves using balanced chemical equations to determine the proportions of substances involved. In the case of KO2 reacting with CO2, stoichiometry helps to quantify the amounts of each reactant and product, which is important for understanding the overall reaction and its implications in practical applications like oxygen masks.

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Textbook Question

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (b) Name Ag₃AsO₄ by analogy to the corresponding compound containing phosphorus in place of arsenic.

Textbook Question

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (c) If it took 25.0 mL of 0.102 M Ag⁺to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Textbook Question

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH₃ were drawn into the acid solution?

Textbook Question

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (b) How many ppm of NH₃ were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.)