Question

Dilute aqueous hydrogen peroxide is used as a bleaching agent and for disinfecting surfaces and small cuts. Its concentration is sometimes given as a certain number of "volumes hydrogen peroxide," which refers to the number of volumes of \(\mathrm{O}_{2}\) gas, measured at STP, that a given volume of hydrogen peroxide solution will release when it decomposes to \(\mathrm{O}_{2}\) and liquid \(\mathrm{H}_{2} \mathrm{O}\). How many grams of hydrogen peroxide are in \(0.100 \mathrm{~L}\) of \(" 20\) volumes hydrogen peroxide" solution?

Short answer

6.07 g of H2O2.

Step-by-step solution

Understanding the Problem

The problem is asking for the amount of hydrogen peroxide in grams in a 0.100 L solution labeled as '20 volumes hydrogen peroxide'. This means the solution releases 20 times its volume in oxygen gas when it decomposes.

Decomposition Reaction of Hydrogen Peroxide

The decomposition reaction of hydrogen peroxide is: \( 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \) This reaction tells us that 2 moles of hydrogen peroxide produce 1 mole of oxygen gas.

Calculating Volume of Oxygen Gas

Given 0.100 L of solution and '20 volumes', the volume of oxygen gas produced is: \( 0.100 \text{ L} \times 20 = 2.00 \text{ L} \)

Converting Volume of Gas to Moles

Using the ideal gas law at STP (1 mole of any gas occupies 22.4 L): \( \text{moles of } \text{O}_2 = \frac{2.00 \text{ L}}{22.4 \text{ L/mol}} = 0.0893 \text{ mol} \)

Using Stoichiometry to Determine Moles of H2O2

From the balanced equation, 2 moles of H2O2 produce 1 mole of O2. Therefore: \( \text{moles of } \text{H}_2\text{O}_2 = 2 \times 0.0893 \text{ mol} = 0.1786 \text{ mol} \)

Calculating Mass of H2O2

The molar mass of H2O2 is 34.01 g/mol. Hence, the mass of hydrogen peroxide is: \( \text{mass of } \text{H}_2\text{O}_2 = 0.1786 \text{ mol} \times 34.01 \text{ g/mol} = 6.07 \text{ g} \)

Key concepts

stoichiometry

Stoichiometry deals with the calculation of reactants and products in chemical reactions. This topic is crucial for predicting the quantities of substances consumed and produced in a reaction. To understand stoichiometry, always start with a balanced chemical equation. In the decomposition of hydrogen peroxide (\text{H}_2\text{O}_2), the balanced equation is: \[2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \]
From the equation, 2 moles of hydrogen peroxide decompose to form 2 moles of water and 1 mole of oxygen gas.

Let's break it down further:

• Identify the molar relationships between reactants and products.
• Use these relationships to convert between moles of different compounds.

In the provided solution, we use this relationship to calculate how much \text{H}_2\text{O}_2 is needed to produce the given amount of oxygen gas. This thorough understanding of stoichiometry ensures we can predict how much chemical substance we need or expect in reactions.

ideal gas law

The Ideal Gas Law is a fundamental principle in chemistry, represented by the equation \(PV = nRT \). This equation relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of an ideal gas. In our exercise, we use it to convert the volume of \text{O}_2 gas to moles because the volume is given and we need the amount in moles to proceed with stoichiometry.

Steps involved:

• Identify the volume of the gas at standard temperature and pressure (STP). For \text{O}_2, this is given as 2.00 L.
• Use the molar volume of a gas at STP (22.4 L/mol) to find the moles of \text{O}_2.
• Apply the conversion: \( \text{moles of} \text{O}_2 = \frac{2.00 \text{ L}}{22.4 \text{ L/mol}} = 0.0893 \text{ mol} \).

This step is crucial because it connects the volume of gas we can measure directly to the amount in moles, which we then use in further chemical calculations.

molar mass calculation

Calculating the molar mass of a compound is essential for translating between moles and grams. The molar mass is the mass of one mole of a substance, expressed in g/mol. To calculate it, sum the atomic masses of all atoms in the molecule.
For hydrogen peroxide (\text{H}_2\text{O}_2), the molar mass calculation is:

• Hydrogen's atomic mass is approximately 1.01 g/mol. Since there are 2 atoms, the total for hydrogen is 2 × 1.01 = 2.02 g/mol.
• Oxygen's atomic mass is approximately 16.00 g/mol. With 2 atoms, the total for oxygen is 2 × 16.00 = 32.00 g/mol.
• Therefore, molar mass of \text{H}_2\text{O}_2 = 2.02 + 32.00 = 34.02 g/mol.

Once you have the molar mass, you can convert between grams and moles using: \( \text{mass} = \text{moles} \times \text{molar mass} \).
In the example: \( \text{mass of}\text{ H}_2\text{O}_2 = 0.1786 \text{ mol} \times 34.01 \text{ g/mol} = 6.07 \text{ g} \).
This step provides the final answer in the exercise, showing the mass of hydrogen peroxide in the solution.