Question

Concentrated hydrogen peroxide solutions are explosively de-composed by traces of transition metal ions (such as \(\mathrm{Mn}\) or \(\mathrm{Fe}\) ): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(g)\), collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

Short answer

The volume of pure \(O_2(g)\), collected at \(27^{\circ}C\) and \(746\text{ torr}\), generated by decomposition of \(125g\) of a \(50.0\%\) by mass hydrogen peroxide solution can be calculated through the following steps: 1. Calculate the moles of hydrogen peroxide in the solution: $$ Moles\,of\,H_2O_2 = \frac{0.50 \times 125\,g}{34.0\,g/mol} $$ 2. Determine moles of oxygen gas generated: $$ Moles\,of\,O_2 = \frac{1}{2} \times Moles\,of\,H_2O_2 $$ 3. Calculate the volume of oxygen gas using the Ideal Gas Law: $$ V = \frac{nRT}{P} = \frac{(Moles\,of\,O_2) \times (0.0821\,L\,atm/mol\,K) \times (300.15\,K)}{\frac{746}{760}\text{ atm}} $$ Compute the values in these equations to find the final volume of oxygen gas generated in liters.

Step-by-step solution

Calculate moles of hydrogen peroxide in the solution

First, find the mass of hydrogen peroxide in the solution: Mass of H₂O₂ = 50% of 125g $$ Mass\,of\,H_2O_2 = 0.50 \times 125\,g $$ Calculate moles of hydrogen peroxide using its molar mass (H₂O₂): 2(1.0) + 2(16.0) = 34.0 g/mol $$ Moles\,of\,H_2O_2 = \frac{Mass\,of\,H_2O_2}{Molar\,mass\,of\,H_2O_2} $$

Determine moles of oxygen gas generated

Next, use stoichiometry to find the moles of oxygen gas generated. The balanced chemical equation is: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ From the equation, 2 moles of hydrogen peroxide (H₂O₂) generates 1 mole of oxygen gas (O₂). Thus, $$ Moles\,of\,O_2 = \frac{1}{2} \times Moles\,of\,H_2O_2 $$

Calculate the volume of oxygen gas

Now we will use the Ideal Gas Law to determine the volume of oxygen gas at 27°C (300.15 K) and 746 torr. Ideal Gas Law: \(PV = nRT\) Convert the pressure from torr to atm: \(1\, atm = 760 \, torr\), hence Pressure = \(\frac{746}{760}\) atm. Given: n (moles of O₂) = from Step 2, R (gas constant) = 0.0821 L atm/mol K, T (temperature in Kelvin) = 27°C + 273.15 = 300.15 K. Find the volume (V) by rearranging the Ideal Gas Law equation: $$ V = \frac{nRT}{P} $$ Calculate the volume of oxygen gas (in liters) generated by the decomposition of the given concentration of hydrogen peroxide.

Key concepts

Chemical Reaction Balancing

Chemical reaction balancing is a fundamental process in stoichiometry that ensures the law of conservation of mass is obeyed in a chemical equation. Every chemical equation represents a recipe for how reactants are transformed into products. To balance a reaction, we need to make sure that the number of atoms of each element on the reactants side equals the number on the products side.

In our provided exercise, the decomposition of hydrogen peroxide (2 H_2 O_2(aq)) into water (H_2 O(l)) and oxygen gas (O_2(g)) is already balanced with the coefficients 2, 2, and 1 respectively. This means for every two molecules of hydrogen peroxide that decompose, two molecules of water and one molecule of oxygen are formed. Balancing chemical equations is a vital step before any stoichiometric calculations can be performed, as it accurately relates the quantities of reactants and products involved.

Molar Mass Calculation

Molar mass calculation is key to converting between grams and moles of a substance, which is essential to stoichiometry. The molar mass of a compound is the weight of one mole of that compound and is expressed in grams per mole (g/mol). It can be calculated by summing the masses of the individual atoms (from the periodic table) in the chemical formula.

In this exercise, the molar mass of hydrogen peroxide (H_2 O_2) was calculated by adding the atomic masses of hydrogen (1.0 g/mol for each of the two atoms) and oxygen (16.0 g/mol for each of the two atoms), resulting in a total of 34.0 g/mol for hydrogen peroxide. Understanding how to find molar mass allows students to move forward in stoichiometric calculations and determine the moles of substances involved in the reaction.

Ideal Gas Law Application

Application of the Ideal Gas Law is crucial when dealing with gases in chemical reactions. The Ideal Gas Law is an equation that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas to the gas constant (R). This law is commonly stated as PV = nRT.

For the calculation in our exercise, the Ideal Gas Law is used to determine the volume of oxygen gas produced. After obtaining the number of moles of the gas, we apply the law with the known values for temperature (in Kelvin) and pressure (converted to atmospheric pressure), along with the gas constant R = 0.0821 L atm/mol K. The calculation involves rearranging the law to solve for V, which gives us the volume of gas produced under the given conditions.

Gas Volume Calculation

Gas volume calculation in stoichiometry involves using derived or known quantities from a balanced chemical equation alongside the ideal gas law. The goal is to find the volume occupied by a gas at certain conditions of temperature and pressure.

In our problem, we are asked to calculate the volume of pure oxygen gas (O_2(g)) formed from the decomposition of a certain mass of hydrogen peroxide solution. Once the moles of oxygen are found by stoichiometric relationships from the balanced equation, the ideal gas law comes into play. We input the moles of oxygen, the pressure in atmospheres, the temperature in Kelvin, and the gas constant to calculate the desired volume. The result is the volume of oxygen gas that would be collected under the specified conditions, representing a practical application of stoichiometry in predicting the outcome of a chemical reaction involving gases.