Question

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 764 torr, would be generated by decomposition of 125 g of a \(50.0 \%\) by mass hydrogen peroxide solution?

Short answer

The volume of pure oxygen gas collected at \(27^{\circ} \mathrm{C}\) and 764 torr generated by the decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution is approximately 22.6 L.

Step-by-step solution

1. Calculate the mass of hydrogen peroxide in the solution

First, we need to determine the mass of hydrogen peroxide (H2O2) in the solution given its percentage by mass. The mass of hydrogen peroxide in the solution can be calculated as follows: Mass of H2O2 = (Percentage by mass / 100) * Total mass of solution Mass of H2O2 = (50.0 / 100) * 125 g Mass of H2O2 = 62.5 g

2. Determine the number of moles of hydrogen peroxide

Next, we need to convert the mass of hydrogen peroxide into moles using its molar mass. The molar mass of hydrogen peroxide (H2O2) is 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol. Number of moles of H2O2 = Mass of H2O2 / Molar mass of H2O2 Number of moles of H2O2 = 62.5 g / 34.02 g/mol Number of moles of H2O2 = 1.84 mol

3. Determine the number of moles of oxygen gas produced

Using the balanced chemical equation, 2 H2O2 (aq) → 2 H2O (l) + O2 (g), we can determine the number of moles of oxygen gas, O2, produced for the decomposition. From the balanced chemical equation, it is clear that 2 moles of hydrogen peroxide produces 1 mole of oxygen gas. Therefore: Number of moles of O2 = 0.5 * Number of moles of H2O2 Number of moles of O2 = 0.5 * 1.84 mol Number of moles of O2 = 0.92 mol

4. Calculate the volume of oxygen gas produced

Now, we will use the ideal gas law to calculate the volume of oxygen gas produced: PV = nRT Here, P = pressure = 764 torr (convert to atm by dividing by 760 torr/atm) = 1.005 atm V = volume (which we want to find) n = number of moles of O2 = 0.92 mol R = ideal gas constant = 0.0821 (L atm) / (mol K) T = temperature = 27 °C (convert to K by adding 273) = 300 K We can rearrange the ideal gas law equation to solve for volume: V = nRT / P Now, we will plug in the values and find the volume of oxygen gas produced: V = (0.92 mol * 0.0821 L atm/(mol K) * 300 K) / 1.005 atm V = 22.6 L

Answer

The volume of pure oxygen gas collected at 27 °C and 764 torr generated by the decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution is approximately 22.6 L.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is key to predicting the outcomes of chemical reactions. Stoichiometry comes from the Greek words \( \text{stoicheion} \) meaning 'element' and \( \text{metron} \) meaning 'measure'. It involves calculating the amounts of reactants and products involved in a chemical reaction.

In the case of hydrogen peroxide decomposition, we use stoichiometry to relate the amount of hydrogen peroxide that reacts to the amount of oxygen gas produced. The balanced chemical equation gives us the necessary proportions: \( 2 \mathrm{H}_{2} \mathrm{O}_{2}(aq) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g) \. \) This tells us that for every two moles of hydrogen peroxide, one mole of oxygen gas is produced.

To solve problems using stoichiometry, we start by converting mass to moles, because moles allow us to use the molar ratios from the balanced equation. These ratios serve as a conversion factor between reactants and products, leading us to step 3 in our original solution. Here, the student can see that the stoichiometric relationship is the core driver of being able to predict how much oxygen gas will be produced from a known quantity of hydrogen peroxide.

Molar Mass Calculation

Molar mass acts as a bridge between the mass of a substance and the number of moles, a fundamental concept used in the step-by-step solution provided. The molar mass of a compound is calculated by summing the masses of all the atoms in the formula. For hydrogen peroxide (\( \mathrm{H}_{2}\mathrm{O}_{2} \)), its molar mass is the sum of the masses of two hydrogens (1.01 g/mol each) and two oxygens (16.00 g/mol each), resulting in 34.02 g/mol.

In our exercise, knowing the molar mass of hydrogen peroxide allows us to convert the given mass (62.5 g) to the number of moles necessary for stoichiometric calculations. The formula for this conversion is simply:
\[ \text{Number of moles} = \frac{\text{Mass of substance}}{\text{Molar mass of substance}} \]
Remember, this crucial step enables students to move from the tangible mass measurements one can make in a lab to the mole concept, which underpins chemical reaction calculations and stoichiometry.

Ideal Gas Law

The ideal gas law is an equation that relates the pressure, volume, temperature, and number of moles of a gas. It's given by \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

In the problem’s context, we use the ideal gas law to find the volume of oxygen gas produced at specific conditions of temperature and pressure. We convert the given temperature to Kelvin and the pressure to atmospheres, using these values to calculate the volume via the rearranged ideal gas law: \(V = \frac{nRT}{P}\).

This relationship makes it possible to determine the expected volume of a gas formed in a reaction, as seen in step 4 of the provided solution. The ideal gas law is a fundamental principle in chemistry and is instrumental in a wide range of applications, from calculating gas volumes in chemical reactions to understanding the behavior of gases under different conditions.