Problem 62 - Chemistry The Molecular Science

Question

Drug stores sell \(3 \%\) aqueous hydrogen peroxide that is used as an antiseptic. Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) decomposes to water and oxygen. Calculate the volume of oxygen produced if \(250 . \mathrm{mL}\) of \(3 \%\) hydrogen peroxide decomposes fully at \(750 . \mathrm{mmHg}\) and \(22^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Volume of \(\text{O}_2\) produced is approximately 2.67 L.

Step 1: Understand the Decomposition Reaction

The decomposition of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) can be represented by the balanced chemical equation: \[2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2.\] This means two moles of hydrogen peroxide decompose to produce one mole of oxygen gas \(\text{O}_2\).

Step 2: Convert Mass Percent to Grams

We are given a solution that is \(3\%\) hydrogen peroxide by mass. For a \(250 \text{ mL}\) solution, the mass of \(\text{H}_2\text{O}_2\) is calculated as follows, assuming the density of the solution is similar to that of water (\(1 \text{ g/mL}\)):\[\text{Mass of solution} = 250 \text{ g}\]\[\text{Mass of } \text{H}_2\text{O}_2 = 0.03 \times 250 = 7.5 \text{ g}.\]

Step 3: Convert Grams to Moles

Use the molar mass of hydrogen peroxide to convert grams to moles. The molar mass of \(\text{H}_2\text{O}_2\) is approximately \(34.01 \text{ g/mol}\):\[\text{Moles of } \text{H}_2\text{O}_2 = \frac{7.5 \text{ g}}{34.01 \text{ g/mol}} \approx 0.2205 \text{ mol}.\]

Step 4: Determine Moles of Oxygen Produced

From the balanced equation, two moles of \(\text{H}_2\text{O}_2\) produce one mole of \(\text{O}_2\). Therefore, the moles of oxygen produced are:\[\text{Moles of } \text{O}_2 = \frac{0.2205}{2} \approx 0.11025 \text{ mol}.\]

Step 5: Use Ideal Gas Law to Calculate Volume

We will use the ideal gas law \(PV = nRT\) to determine the volume of \(\text{O}_2\), using \(n = 0.11025 \text{ mol}\), pressure \(P = 750 \text{ mmHg} = 0.9869 \text{ atm}\), and temperature \(T = 22^\circ\text{C} = 295 \text{ K}\).The ideal gas constant \(R\) is \(0.0821 \text{ L atm/mol K}\):\[0.9869 \cdot V = 0.11025 \cdot 0.0821 \cdot 295\]\[V = \frac{0.11025 \cdot 0.0821 \cdot 295}{0.9869} \approx 2.67 \text{ L}.\]

Step 6: Final Volume of Oxygen Produced

The calculated volume of oxygen gas produced from the complete decomposition of the hydrogen peroxide is \(2.67 \text{ L}\) when measured at \(750 \text{ mmHg}\) and \(22^\circ\text{C}\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a vital tool in understanding the behavior of gases. It's expressed by the formula: \(PV = nRT\). Each variable represents a fundamental property of the gas:

• \(P\) stands for pressure, typically measured in atm (atmospheres) or mmHg.
• \(V\) is the volume of the gas, measured in liters.
• \(n\) represents the number of moles, a measure of the amount of substance.
• \(R\) is the ideal gas constant, which is 0.0821 L atm/mol K.
• \(T\) is the temperature, always expressed in Kelvin (K).

In our problem, we use the Ideal Gas Law to find the volume (V) of oxygen gas produced after decomposition. We need the number of moles \(n\), the pressure \(P\) in atm, and the temperature \(T\) in Kelvin. Filling these into the equation gives us the desired volume.

Chemical Equation

A chemical equation describes what happens in a chemical reaction. It uses chemical formulas to show which substances change into which other substances. For hydrogen peroxide (\(\text{H}_2\text{O}_2\)), the decomposition is represented as:\[2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2.\]This balanced equation lets us see how many moles of each substance are involved.

• It indicates that 2 moles of hydrogen peroxide decompose to form 2 moles of water (\(\text{H}_2\text{O}\)) and 1 mole of oxygen gas (\(\text{O}_2\)).

Understanding this helps us calculate the moles of oxygen produced, which is critical for using the Ideal Gas Law.

Moles Calculation

In chemistry, a mole is a unit that measures the amount of a substance. It helps us quantify how much of each reactant or product we have. To calculate moles, we use the molar mass, the mass of one mole of a substance.Given the decomposition reaction of hydrogen peroxide, we first calculate how many grams of \(\text{H}_2\text{O}_2\) are in the solution. We know that 3% of the 250 mL solution is hydrogen peroxide. By multiplying, we find:\[\text{Mass of } \text{H}_2\text{O}_2 = 0.03 \times 250 = 7.5 \text{ g}.\]Next, we use the molar mass to convert grams to moles:\[\text{Molar mass of } \text{H}_2\text{O}_2 = 34.01 \text{ g/mol}.\]\[\text{Moles of } \text{H}_2\text{O}_2 = \frac{7.5}{34.01} \approx 0.2205 \text{ mol}.\]This conversion is essential for finding the moles of oxygen produced.

Molar Mass

Molar mass is the mass of one mole of a substance and is crucial for mole calculations. It is given in units of g/mol (grams per mole). Each compound has a molar mass, which can be found by adding the atomic masses of all atoms in the molecule. For hydrogen peroxide (\(\text{H}_2\text{O}_2\)), the molar mass calculation is as follows:

• Hydrogen: 2 atoms \(\times\) 1.01 g/mol = 2.02 g/mol
• Oxygen: 2 atoms \(\times\) 16.00 g/mol = 32.00 g/mol

Adding these gives us a molar mass of 34.01 g/mol.Understanding molar mass allows us to convert the mass of a substance to moles, enabling us to predict how much product will form in a reaction. It's the bridge between the world of chemistry and the measureable real-world quantities.