Question

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is used to disinfect contact lenses. How many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) can be liberated from \(10.0 \mathrm{mL}\) of an aqueous solution containing \(3.00 \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass? The density of the aqueous solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(1.01 \mathrm{g} / \mathrm{mL}\) $$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{O}_{2}(\mathrm{g})$$

Short answer

Therefore, 111 mL of \(O_{2}\) gas can be liberated from 10.0 mL of an aqueous solution containing 3.00% \(H_{2}O_{2}\) by mass.

Step-by-step solution

Calculate the mass of \(H_{2}O_{2}\)

First, find the mass fraction of \(H_{2}O_{2}\) in the solution. The solution is made up of 3.00% \(H_{2}O_{2}\) by mass. Therefore, in 10.0 mL of the solution, the mass of \(H_{2}O_{2}\) is \(10.0 \, mL * 1.01 \, g/mL * 0.03 = 0.303 \, g\).

Convert mass of \(H_{2}O_{2}\) to moles

The molar mass of \(H_{2}O_{2}\) is \(16*2+1*2 = 34 \, g/mol\). So, we have \(0.303 \, g / 34 \, g/mol = 0.00892 \, mol\) of \(H_{2}O_{2}\).

Determine the amount of \(O_{2}\) liberated

From the balanced equation, 2 moles of \(H_{2}O_{2}\) gives 1 mole of \(O_{2}\). Therefore, \(0.00892 \, mol * (1/2) = 0.00446 \, mol\) of \(O_{2}\) is produced.

Apply the Ideal Gas Law

Finally, use the ideal gas law, \(PV=nRT\), where \(P=752 \, mmHg = 752/760 \, atm = 0.990 \, atm\), \(R=0.0821 \, atm.L/mol.K\), and \(T=22°C=22+273=295 \, K\). Solving for \(V\), we have \(V=nRT/P = 0.00446 \, mol * 0.0821 \, L/mol.K * 295 \, K / 0.990 \, atm = 0.111 \, L = 111 \, mL\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a powerful equation used to relate the behavior of gases under various conditions. It combines several important gas properties into one equation and is expressed as: \[ PV = nRT \]- **P** is the pressure of the gas, usually in atmospheres.- **V** is the volume of the gas, in liters.- **n** stands for the number of moles of gas.- **R** is the ideal gas constant, approximately 0.0821 L·atm/mol·K.- **T** is the temperature in Kelvin.
When using the Ideal Gas Law, it’s essential to ensure all units are compatible. For instance, converting pressure from mmHg to atm, or temperature from Celsius to Kelvin. This step is crucial for accurate calculations.
In the exercise, we determined the volume of oxygen (\(O_{2}\)) gas released by applying this formula. It ties together the mathematical relationship between the volume generated and the conditions of temperature and pressure under which the gas is released.

Moles and Molar Mass

Understanding moles and molar mass is key to grasping stoichiometry, which deals with the calculation of reactants and products in a chemical reaction. - **Moles** measure quantity in chemistry. One mole equals Avogadro’s number, or approximately \(6.022 \times 10^{23}\) molecules or atoms.- **Molar Mass** is the mass of one mole of a substance, measured in g/mol.
Converting mass to moles allows you to traverse from the mass of a substance to the number of particles. Knowing the molar mass of each element in a compound can help calculate the compound’s molar mass.
In the example exercise, we converted the mass of hydrogen peroxide (\(H_{2}O_{2}\)) to moles using its molar mass. This conversion forms the basis for determining how much of another product, like \(O_{2}\), is created in the reaction.

Chemical Reactions and Equations

Chemical reactions describe a process where substances transform into different substances. Balanced chemical equations give a clear depiction of these transformations. - **Reactants** are substances consumed in the reaction.- **Products** are substances formed in the reaction.- **Coefficients** in the equation indicate the ratio of moles of each substance.
A balanced chemical reaction ensures that the number of atoms of each element is conserved. This follows the Law of Conservation of Mass, stating that matter is neither created nor destroyed.
In our exercise, the equation \(2 \, H_{2}O_{2} \rightarrow 2 \, H_{2}O + O_{2}\) was critical for understanding how much oxygen gas (\(O_{2}\)) is produced from hydrogen peroxide. By using the stoichiometric coefficients, we converted moles of \(H_{2}O_{2}\) to moles of \(O_{2}\).