Question

Hydrogen peroxide with a concentration of \(3.0 \%\) (3.0 g of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in \(100 \mathrm{~mL}\) of solution) is sold in drugstores for use as an antiseptic. For a \(10.0-\mathrm{mL} 3.0 \% \mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate (a) the oxygen gas produced (in liters) at STP when the compound undergoes complete decomposition and (b) the ratio of the volume of \(\mathrm{O}_{2}\) collected to the initial volume of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

Short answer

0.0988 L of \( \mathrm{O}_{2} \) produced; the volume ratio is 9.88.

Step-by-step solution

Understanding the Decomposition Reaction

Hydrogen peroxide ( \( \mathrm{H}_{2} \mathrm{O}_{2} \) ) decomposes according to the reaction:\[2 \mathrm{H}_{2} \mathrm{O}_{2} (l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g)\]For every 2 moles of \( \mathrm{H}_{2} \mathrm{O}_{2} \), 1 mole of \( \mathrm{O}_{2} \) is produced.

Calculate Moles of Hydrogen Peroxide

We start by finding the moles of \( \mathrm{H}_{2} \mathrm{O}_{2} \) in the solution. Given that we have a\( 3.0 \% \) solution, there are 3.0 g of \( \mathrm{H}_{2} \mathrm{O}_{2} \) in every 100 mL. For a 10.0 mL solution, the grams of \( \mathrm{H}_{2} \mathrm{O}_{2} \) are:\[\frac{10.0 \text{ mL}}{100 \text{ mL}} \times 3.0 \text{ g} = 0.3 \text{ g}\]Next, convert grams to moles using the molar mass of \( \mathrm{H}_{2} \mathrm{O}_{2} \), which is 34.0 g/mol:\[0.3 \text{ g} \times \frac{1 \text{ mol}}{34.0 \text{ g}} = 0.00882 \text{ mol}\]

Calculate Moles of Oxygen Gas

From the decomposition reaction, 1 mole of \( \mathrm{O}_{2} \) is produced for every 2 moles of \( \mathrm{H}_{2} \mathrm{O}_{2} \). Therefore, the moles of \( \mathrm{O}_{2} \) produced is:\[\frac{0.00882 \text{ mol}}{2} = 0.00441 \text{ mol} \mathrm{O}_{2}\]

Convert Moles of Oxygen to Volume at STP

At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. So, the volume of \( \mathrm{O}_{2} \) produced is:\[0.00441 \text{ mol} \times 22.4 \text{ L/mol} = 0.0988 \text{ L}\]

Calculate Volume Ratio

Finally, calculate the ratio of the volume of \( \mathrm{O}_{2} \) collected to the initial volume of the \( \mathrm{H}_{2} \mathrm{O}_{2} \) solution:\[\frac{0.0988 \text{ L}}{0.010 \text{ L}} = 9.88\]This means the volume of \( \mathrm{O}_{2} \) is approximately 9.88 times the initial volume of the solution.

Key concepts

Oxygen Gas Production

Hydrogen peroxide (\( \mathrm{H}_{2} \mathrm{O}_{2} \)) is a simple chemical compound that decomposes when it breaks down, often with the help of a catalyst or by itself over time, into water and oxygen gas. This decomposition reaction can be represented by the equation: \[2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g)\] Here, two molecules (or moles) of hydrogen peroxide decompose to produce one molecule (or mole) of oxygen gas and two molecules (or moles) of water. The oxygen gas produced in this reaction is often used as an example in labs to demonstrate gas production from a liquid source. In practical scenarios, this decomposition has significance in various applications, from generating oxygen for medical uses to catalyzing reactions in industrial processes.

Moles Calculation

The first step in calculations involving chemical reactions is often to determine the amount of a substance in moles. A mole is a standard unit in chemistry representing approximately \(6.022 \times 10^{23}\) particles of the substance.To calculate moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\), we first need to convert the given mass of hydrogen peroxide into moles using its molar mass. The provided hydrogen peroxide solution contains \(3.0\%\) by mass, meaning that 3 grams of \(\mathrm{H}_{2} \mathrm{O}_{2}\) are present in every 100 mL of solution. Therefore, for 10.0 mL, we have:

• \(0.3 \text{ g of } \mathrm{H}_{2} \mathrm{O}_{2} \)
• Molar Mass of \( \mathrm{H}_{2} \mathrm{O}_{2} = 34.0 \text{ g/mol} \)
• \(\text{Number of moles} = \frac{0.3 \text{ g}}{34.0 \text{ g/mol}} = 0.00882 \text{ moles} \)

Using this calculated amount of moles is critical as it allows you to determine how many moles of other substances will be involved, such as the production of gas, in the chemical reaction.

Volume Ratio

In many chemical reactions, comparing the volumes of reactants and products can offer insights into the reaction's efficiency and yield. For the decomposition of hydrogen peroxide, the comparison is made between the volume of produced oxygen and the initial volume of the hydrogen peroxide solution.Firstly, we calculate how much oxygen gas is produced using the moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) decomposed. Then, we find the volume of this oxygen gas at standard conditions of temperature and pressure (STP). Finally, we can find the volume ratio:

• Volume of \( \mathrm{O}_{2} \): Calculated using moles and known volume per mole at STP
• Initial Volume of \( \mathrm{H}_{2} \mathrm{O}_{2} \): 10.0 mL converted to liters
• Volume Ratio: \( \frac{\text{Volume of } \mathrm{O}_{2}}{\text{Volume of } \mathrm{H}_{2} \mathrm{O}_{2}} = \frac{0.0988 \text{ L}}{0.010 \text{ L}} = 9.88 \)

This shows that the produced oxygen has a much larger volume than the liquid hydrogen peroxide it came from, which demonstrates the expansion gas undergoes from a liquid reactant.

Standard Temperature and Pressure (STP)

STP, or Standard Temperature and Pressure, refers to the conditions under which the behavior of gases is often described: a temperature of 0 degrees Celsius (273.15 K) and a pressure of 1 atm (atmosphere). Under these conditions, gases exhibit predictable behaviors, such as the volume one mole of gas will occupy. One of the key aspects of STP is that one mole of any ideal gas will occupy a volume of 22.4 liters. This concept is widely used in chemical calculations, such as to easily convert between moles of gas and volume:

• It simplifies conversions during chemical experiments.
• Provides a common baseline for scientists worldwide.
• Essential for calculating the expected gas volume produced or needed in reactions.

Understanding STP is crucial for accuracy when predicting outcomes in experiments, especially those involving gases.