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Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$ \mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q) $$ What mass of hydrogen peroxide should result when \(1.50 \mathrm{~g}\) barium peroxide is treated with \(25.0 \mathrm{~mL}\) hydrochloric acid solution containing \(0.0272 \mathrm{~g} \mathrm{HCl}\) per \(\mathrm{mL}\) ? What mass of which reagent is left unreacted?

Short Answer

Expert verified
The mass of hydrogen peroxide formed is \(1.52\,g\) and the unreacted mass of hydrochloric acid is \(0.0324\,g\).

Step by step solution

01

Calculate moles of barium peroxide and hydrochloric acid

We need to first convert the mass of barium peroxide and the volume of hydrochloric acid into moles. To do this, divide the mass by the molar mass for each substance: Moles of barium peroxide: \( moles_{BaO2} = \frac{mass_{BaO2}}{molar\,mass_{BaO2}}=\frac{1.50\,g}{(137.327+2×16)\,g/mol}=\frac{1.50\,g}{169.327\,g/mol} = 0.008864\,mol \) Moles of hydrochloric acid: \( moles_{HCl}=volume_{HCl}\times concentration_{HCl}=25.0\,mL\times\frac{0.0272\,g}{mL}\times\frac{1\,mol}{36.461\,g}= 0.01862\,mol \)
02

Identify the limiting reagent

The balanced equation shows that 1 mole of barium peroxide reacts with 2 moles of hydrochloric acid. To determine the limiting reagent, we need to compare the mole ratios: \( \frac{moles_{BaO2}}{moles_{HCl}} = \frac{0.008864\,mol}{0.01862\,mol} = 0.4761 < 0.5 \) Since the ratio is less than 0.5, Barium peroxide is the limiting reagent.
03

Calculate the moles of hydrogen peroxide formed

Now we can use the balanced equation to calculate moles of hydrogen peroxide produced: \( moles_{H2O2} = moles_{BaO2} \times \frac{1\,mol\,H2O2}{1\,mol\,BaO2} = 0.008864\,mol \)
04

Convert moles of hydrogen peroxide to mass

Next, convert the moles of hydrogen peroxide to mass by multiplying with its molar mass: \( mass_{H2O2} = moles_{H2O2}\times molar\,mass_{H2O2} = 0.008864\,mol\times(2×1.00784+2×16)\,g/mol = 1.52\,g \)
05

Calculate the mass of unreacted reagent

Since barium peroxide is the limiting reagent, all of it reacts. To find the unreacted hydrochloric acid, we need to calculate how much of it reacted and then subtract it from the initial amount: Moles of reacted hydrochloric acid: \( moles_{reacted\,HCl} = moles_{BaO2} \times \frac{2\,mol\,HCl}{1\,mol\,BaO2} = 0.008864\,mol\times 2 = 0.01773\,mol \) Mass of unreacted hydrochloric acid: \( mass_{unreacted\,HCl} = (moles_{HCl} - moles_{reacted\,HCl})\times molar\,mass_{HCl} \) \( mass_{unreacted\,HCl} = (0.01862\,mol - 0.01773\,mol)\times 36.461\,g/mol = 0.0324\,g \) So the mass of hydrogen peroxide formed is \(1.52\,g\) and the unreacted mass of hydrochloric acid is \(0.0324\,g\).

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Key Concepts

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

Limiting Reagent
In chemistry, the concept of the limiting reagent is quite straightforward yet essential for understanding chemical reactions. When two or more reactants are combined, they react, forming products until one of the reactants is entirely used up. The reactant that is completely consumed first is known as the limiting reagent. It determines the maximum amount of product that can be formed. The other reactants, which are not entirely used, are referred to as excess reagents.
To identify the limiting reagent, one must use stoichiometry, which involves calculating the moles of each reactant and comparing them according to the balanced chemical equation. For instance, in the reaction between barium peroxide and hydrochloric acid to produce hydrogen peroxide, one mole of barium peroxide reacts with two moles of hydrochloric acid. Thus, by calculating and comparing their molar ratios, we deduce that barium peroxide is the limiting reagent.
Chemical Reactions
Chemical reactions are processes whereby reactants transform into products. They involve the breaking and forming of chemical bonds, and they often result in the conversion of substances into entirely new compounds, possessing different properties from the original materials. The descriptions of chemical reactions are represented through chemical equations.
In the case of hydrogen peroxide formation from barium peroxide and hydrochloric acid, the chemical equation is: \[ \mathrm{BaO}_{2}(s) + 2 \mathrm{HCl}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(aq) + \mathrm{BaCl}_{2}(aq) \]. This balanced equation signifies that the reactants and products are in the correct stoichiometric proportions, ensuring the conversation of mass. These reactions might also involve energy changes, such as heat release or absorption, and physical changes, such as gas formation, like oxygen in this case.
Oxidizing Agent
An oxidizing agent is a substance that gains electrons in a chemical reaction, causing another substance to lose electrons, or oxidize. In the context of reactions, oxidizing agents are crucial because they facilitate the transfer of electrons between substances.
Hydrogen peroxide is an excellent example of an oxidizing agent. It is commonly used because of its ability to effectively kill bacteria and other pathogens by releasing oxygen. This property makes hydrogen peroxide especially useful in medical settings, such as cleaning wounds and cuts. It decomposes into water and oxygen, making it safer and environmentally friendly after use.
  • It is involved in various oxidation reactions, due to its ability to donate oxygen.
  • In the discussed reaction, hydrogen peroxide is the product but can act as an oxidizing agent in different scenarios.
Therefore, understanding oxidizing agents helps explain why certain substances might act as effective reagents in medical and technical applications.

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Most popular questions from this chapter

Maleic acid is an organic compound composed of \(41.39 \% \mathrm{C}\), \(3.47 \% \mathrm{H}\), and the rest oxygen. If \(0.129 \mathrm{~mol}\) of maleic acid has a mass of \(15.0 \mathrm{~g}\), what are the empirical and molecular formulas of maleic acid?

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right) .\) Calculate the percent composition by mass of fluoxymesterone.

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$ \begin{aligned} 3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) & \longrightarrow \\ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of \(\overline{\mathrm{Al}}\) ?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

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