Concept explainers
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:
BaO2(s) + 2HCl(aq)→H2O2(aq) + BaCl2(aq)
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: 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,
Mass of the substance=(Number of moles)×(Molar mass of the substance)
To determine: The mass of hydrogen peroxide that can be produced by the given chemical reaction and the mass of the unreacted reagent.
Answer to Problem 119E
The amount of hydrogen peroxide that can be produced is 0.301 g_. The mass of the unreacted reagent, that is HCl, is 3.6×10-2 g_.
Explanation of Solution
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 1.50 g.
The given volume of HCl is 25.0 ml.
The solution contains 0.0272 g HCl per ml.
The molar mass of BaO2 =Ba+2O=(137+(2×16)) g/mol=169 g/mol
Formula
The number of moles of a substance is calculated by the formula,
Number of moles=Given mass of the substanceMolar mass of the substance
Substitute the value of the given mass and the molar mass of BaO2 in the above expression.
Number of moles of BaO2=1.50 g169 g/mol=8.87×10−3 mol
The solution contains 0.0272 g HCl per ml.
Therefore, the mass of HCl =(0.0272×25.0) g=0.68 g
The molar mass of HCl =H+Cl=(1+35.5) g/mol=36.5 g/mol
Number of moles of HCl=0.68 g36.5 g/mol=1.86×10−2 mol
According to the stated reaction,
1 mol of BaO2 reacts with 2 mol of HCl.
8.87×10−3 mol of BaO2 reacts with HCl =(2×8.87×10−3) mol=1.77×10−2 mol
The moles of HCl given are 1.86×10−2 mol.
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,
1 mol of BaO2 produces 1 mol of H2O2.
8.87×10−3 mol of BaO2 produces H2O2 =(1×8.87×10−3) mol=8.87×10−3 mol
Molar mass of H2O2 =2O+2H=((2×16)+(2×1)) g/mol=34 g/mol
The mass of a substance is calculated by the formula,
Mass of the substance=(Number of moles)×(Molar mass of the substance)
Substitute the value of the number of moles of H2O2 and the molar mass of H2O2 in the above expression.
Mass of H2O2=(8.87×10−3 mol)×(34 g/mol)=0.301 g_
To determine: The unreacted amount of the starting material.
Given
The stated chemical reaction is,
BaO2(s)+2HCl(aq)→H2O2(aq)+BaCl2(aq)
The given mass of BaO2 is 1.50 g.
The given volume of HCl is 25.0 ml.
The solution contains 0.0272 g HCl per ml.
The molar mass of BaO2 =Ba+2O=(137+(2×16)) g/mol=169 g/mol
Formula
The number of moles of a substance is calculated by the formula,
Number of moles=Given mass of the substanceMolar mass of the substance
Substitute the value of the given mass and the molar mass of BaO2 in the above expression.
Number of moles of BaO2=1.50 g169 g/mol=8.87×10−3 mol
According to the stated reaction,
1 mol of BaO2 reacts with 2 mol of HCl.
8.87×10−3 mol of BaO2 reacts with HCl =(2×8.87×10−3) mol=1.77×10−2 mol
The moles of HCl given are 1.86×10−2 mol.
Therefore, some amount of HCl is left unreacted. It is the excess reactant.
The moles of HCl left unreacted are calculated by the formula,
Unreacted moles of HCl=Given moles of HCl−Reacted moles of HCl
Substitute the values of the given moles of HCl and the reacted moles of HCl in the above expression.
Unreacted moles of HCl=(1.86×10−2−1.77×10−2) mol=0.09×10−2 mol
Molar mass of HCl =H+Cl=(1+35.5) g/mol=36.5 g/mol
The mass of a substance is calculated by the formula,
Mass of the substance=(Number of moles)×(Molar mass of the substance)
Substitute the value of the number of moles of HCl and the molar mass of HCl in the above expression.
Mass of HCl=(0.09×10−2 mol)×(36.5 g/mol)=3.3×10-2 g_
The amount of hydrogen peroxide that can be produced is 0.301 g_.
The unreacted reagent is HCl and the unreacted mass of HCl is 3.6×10-2 g_.
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Chapter 3 Solutions
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