   Chapter 3, Problem 125E

Chapter
Section
Textbook Problem

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: BaO 2 ( s )   +   2 HCl ( a q ) → H 2 O 2 ( a q )   +  BaCl 2 ( a q ) What mass of hydrogen peroxide should result when 1.50 g barium peroxide is treated with 25.0 mL hydrochloric acid solution containing 0.0272 g HCI per mL? What mass of which reagent is left unreacted?

Interpretation Introduction

Interpretation: The amount of hydrogen peroxide that can be produced by the given chemical reaction is to be calculated. The mass of the unreacted reagent is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

Massofthesubstance=(Numberofmoles)×(Molarmassofthesubstance)

To determine: The mass of hydrogen peroxide that can be produced by the given chemical reaction and the mass of the unreacted reagent.

Explanation

To determine: The mass of hydrogen peroxide that can be produced by the given chemical reaction

Given

The stated chemical reaction is,

BaO2(s)+2HCl(aq)H2O2(aq)+BaCl2(aq)

The given mass of BaO2 is 1.50g .

The given volume of HCl is 25.0ml .

The solution contains 0.0272gHClperml .

The molar mass of BaO2 =Ba+2O=(137+(2×16))g/mol=169g/mol

Formula

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassofthesubstanceMolarmassofthesubstance

Substitute the value of the given mass and the molar mass of BaO2 in the above expression.

NumberofmolesofBaO2=1.50g169g/mol=8.87×103mol

The solution contains 0.0272gHClperml .

Therefore, the mass of HCl =(0.0272×25.0)g=0.68g

The molar mass of HCl =H+Cl=(1+35.5)g/mol=36.5g/mol

NumberofmolesofHCl=0.68g36.5g/mol=1.86×102mol

According to the stated reaction,

1mol of BaO2 reacts with 2mol of HCl .

8.87×103mol of BaO2 reacts with HCl =(2×8.87×103)mol=1.77×102mol

The moles of HCl given are 1.86×102mol .

Therefore, some amount of HCl is left unreacted. It is the excess reactant.

Barium oxide (BaO2) is the limiting reagent.

According to the stated reaction,

1mol of BaO2 produces 1mol of H2O2 .

8.87×103mol of BaO2 produces H2O2 =(1×8.87×103)mol=8.87×103mol

Molar mass of H2O2 =2O+2H=((2×16)+(2×1))g/mol=34g/mol

The mass of a substance is calculated by the formula,

Massofthesubstance=(Numberofmoles)×(Molarmassofthesubstance)

Substitute the value of the number of moles of H2O2 and the molar mass of H2O2 in the above expression

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