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Chapter 9: Problem 5

The volume strength of \(1 \mathrm{M} \mathrm{H}_{2} \mathrm{O}_{2}\) is : (Molar mass of \(\mathrm{H}_{2} \mathrm{O}_{2}=34 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(5.6\) (b) \(16.8\) (c) \(11.35\) (d) \(22.4\)

Short Answer

Expert verified
The volume strength is 11.2; none of the provided options exactly match.

Step by step solution

01

Understanding Volume Strength

The volume strength of hydrogen peroxide ( H鈧侽鈧�) solution refers to the volume of oxygen gas liberated at standard temperature and pressure (STP) by 1 liter of the hydrogen peroxide solution. It tells us how much oxygen is released when the solution decomposes.
02

Decomposition Reaction of H鈧侽鈧�

The decomposition reaction of hydrogen peroxide is:\[2 \mathrm{H}_2\mathrm{O}_2 (aq) \rightarrow 2 \mathrm{H}_2\mathrm{O} (l) + \mathrm{O}_2 (g)\]From this reaction, 2 moles of H鈧侽鈧� decompose to produce 1 mole of O鈧� gas.
03

Calculate Moles of H鈧侽鈧� in 1 Liter

Since we have a 1 M solution, it means there is 1 mole of H鈧侽鈧� in 1 liter of the solution.
04

Calculate Moles of O鈧� Liberated

According to the balanced equation, 2 moles of H鈧侽鈧� produce 1 mole of O鈧�. Therefore, 1 mole of H鈧侽鈧� will produce \( \frac{1}{2} \) mole of O鈧�.
05

Calculate Volume of O鈧� at STP

At STP, 1 mole of a gas occupies 22.4 liters. Therefore, \( \frac{1}{2} \) mole of O鈧� occupies:\[\frac{1}{2} \times 22.4 = 11.2 \text{ liters}\]Hence, the volume strength of 1 M hydrogen peroxide is 11.2.

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Key Concepts

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

Decomposition Reaction of Hydrogen Peroxide
The decomposition reaction of hydrogen peroxide is a straightforward process. In this reaction, hydrogen peroxide (\(\text{H}_2\text{O}_2\)) breaks down into water (\(\text{H}_2\text{O}\)) and oxygen gas (\(\text{O}_2\)). The balanced chemical equation for this reaction is:
\[2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2\]
This means that 2 moles of hydrogen peroxide decompose to produce 2 moles of water and 1 mole of oxygen gas. It is important to fully balance the equation to ensure the number of atoms is consistent on both sides.
  • 1 mole of oxygen gas is generated from 2 moles of H鈧侽鈧�.
  • The reaction is a fundamental example of a decomposition reaction, where a single compound breaks down into simpler substances.
Recognizing the stoichiometry of this reaction is crucial for understanding the volume strength of hydrogen peroxide.
Calculation of Moles
Understanding the concept of moles is key to chemistry. A mole is a unit that measures the amount of substance. One mole contains exactly 6.022 \(\times 10^{23}\) entities, be it atoms, molecules, or ions.

When dealing with a solution, the molarity indicates the number of moles of solute per liter of solution. For a 1 molar (\(1 \, \text{M}\)) hydrogen peroxide solution, this means:
  • 1 mole of \(\text{H}_2\text{O}_2\) is present in 1 liter of solution.
Using the decomposition equation, \(2 \, \text{H}_2\text{O}_2 \rightarrow \text{O}_2\), we see that:
  • 2 moles of \(\text{H}_2\text{O}_2\) produce 1 mole of \(\text{O}_2\)
  • Therefore, 1 mole of \(\text{H}_2\text{O}_2\) will yield \(\frac{1}{2}\) a mole of \(\text{O}_2\)
Grasping how moles relate to the chemical reaction is essential for further calculations.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure, abbreviated as STP, is a term in chemistry that provides a reference framework so that scientists can accurately compare different sets of data. It is defined as:
  • Temperature: 0掳C or 273.15 K
  • Pressure: 1 atm or 101.325 kPa
Many gas laws and calculations, such as the volume of gases, are referenced to STP.
At STP conditions, 1 mole of a gas occupies 22.4 liters of volume. This relationship is important in our case because it allows us to determine the volume strength of hydrogen peroxide. For instance:
  • If \(\frac{1}{2}\) mole of \(\text{O}_2\) is liberated, it will occupy a volume of:\[\frac{1}{2} \times 22.4 \text{ L} = 11.2 \text{ L}\]
This calculation provides insight into the practical implications of gas volume in reactions under standard conditions.
Molar Mass
Molar mass is a critical concept that links the mass of a substance to the amount of substance in moles. It is numerically equal to the average mass of one mole of a substance and is expressed in grams per mole (\(\text{g/mol}\)).
For hydrogen peroxide (\(\text{H}_2\text{O}_2\)), we calculate its molar mass as follows:
  • Hydrogen (\(\text{H}\)) has an atomic mass of 1 gram/mole.
  • Oxygen (\(\text{O}\)) has an atomic mass of 16 grams/mole.
  • Thus, \(\text{H}_2\text{O}_2\) has a molar mass of 2(1) + 2(16) = 34 grams/mole.
Using the molar mass, one can convert the mass of a substance to moles and vice versa, an essential step in stoichiometric calculations. In the context of the volume strength of hydrogen peroxide, the molar mass helps quantify how much \(\text{H}_2\text{O}_2\) is present in a given solution, further assisting in calculating the oxygen generated.

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Most popular questions from this chapter

Hydrogen peroxide, in the pure state, is : (a) non-planar and almost colorless (b) linear and blue in color (c) linear and almost colorless (d) planar and blue in color

The temporary hardness of a water sample is due to compound \(\mathrm{X}\). Boiling this sample converts \(\mathrm{X}\) to compound Y. \(\mathrm{X}\) and \(\mathrm{Y}\), respectively, are : [Main April 12, 2019 (II)] (a) \(\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) (b) \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}\) and \(\mathrm{MgCO}_{3}\) (d) \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)\), and \(\mathrm{CaO}\)

The equation that represents the water-gas shift reaction is: (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \frac{1270 \mathrm{~K}}{\mathrm{Ni}} \rightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(2 \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{~N}_{2}(\mathrm{~g}) \stackrel{1273 \mathrm{~K}}{\longrightarrow}\) \(2 \mathrm{CO}(\mathrm{g})+4 \mathrm{~N}_{2}(\mathrm{~g})\) (c) \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{1270 \mathrm{~K}}{\longrightarrow} \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (d) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \frac{673 \mathrm{~K}}{\text { Catalyst }} \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\)

The one that is NOT suitable for the removal of permanent hardness of water is : [Main Sep. 05, 2020 (II)] (a) Clark's method (b) Ion-exchange method (c) Calgon's method (d) Treatment with sodium carbonate

The numbers of protons, electrons and neutrons in a molecule of heavy water are respectively: [Main Online April 23, 2013] (a) \(8,10,11\) (b) \(10,10,10\) (c) \(10,11,10\) (d) \(11,10,10\)
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