Question

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(i)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathbf{O}_{2}(g),\) 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 \(O_{2}(g)\) generated by the decomposition of 125 g of the 50.0% hydrogen peroxide solution under the given conditions is 22.958 L.

Step-by-step solution

Calculate the amount (in moles) of hydrogen peroxide

We are given 125 g of a 50.0% hydrogen peroxide solution. First, we'll find the mass of hydrogen peroxide in the solution: Mass of hydrogen peroxide = 125 g x 50.0% = 62.5 g Now, we need to find the number of moles of hydrogen peroxide, which can be found by dividing the mass by the molar mass. The molar mass of hydrogen peroxide (H2O2) is 34.02 g/mol (2 x 1.01g/mol for H + 2 x 16.00 g/mol for O): Moles of hydrogen peroxide = (62.5 g)/ (34.02 g/mol) = 1.837 moles

Calculate the amount (in moles) of oxygen gas generated

We can use the stoichiometry of the balanced chemical equation to find the amount of generated O2: $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(i)+\mathrm{O}_{2}(g)$$ From the balanced equation, 2 moles of hydrogen peroxide decompose to produce 1 mole of oxygen gas. Therefore: Moles of oxygen gas generated = (1.837 moles H2O2) x (1 mol O2 / 2 moles H2O2) = 0.9185 moles O2

Calculate the volume of oxygen gas at the given conditions

We are given the following conditions: Temperature (T) = 27°C = 300 K (as we need the temperature in Kelvin for the ideal gas law) Pressure (P) = 746 torr = 0.9818 atm (1 atm = 760 torr) Now we will use the ideal gas law to find the volume (V). The gas constant (R) in the ideal gas law equation is 0.0821 L atm/mol K. Rearrange the ideal gas law for volume and plug in the values: V = nRT/P V = (0.9185 moles O2)(0.0821 L atm/mol K)(300 K)/ (0.9818 atm) Solve for the volume of oxygen gas: V = 22.958 L (rounded to three decimal places) Solution: The volume of pure O2(g) generated by the decomposition of 125 g of the 50.0% hydrogen peroxide solution under the given conditions is 22.958 L.

Key concepts

Stoichiometry Calculation

Understanding stoichiometry is crucial for predicting the outcomes of chemical reactions and for quantifying the relationships between reactants and products. In the decomposition of hydrogen peroxide, stoichiometry helps us determine the amount of oxygen gas produced from a given amount of hydrogen peroxide.

First, we assess the chemical equation: \[2 \mathrm{H}_{2}\mathrm{O}_{2}(aq) \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g)\]. This balanced equation shows that two moles of hydrogen peroxide yield one mole of oxygen gas. Using the mass of hydrogen peroxide and its molar mass, we calculate the number of moles present. We then apply the stoichiometric ratio from the equation to find the corresponding moles of oxygen produced. It's a direct application of the stoichiometric coefficients, showcasing the law of conservation of mass. Always ensure to balance the chemical equation accurately and use the ratio of coefficients to relate moles of reactants to moles of products.

Ideal Gas Law Application

The ideal gas law is a fundamental equation in chemistry that relates the four variables of a gas: pressure (P), volume (V), the number of moles (n), and temperature (T). Expressed as \[PV = nRT\], where 'R' is the universal gas constant.In our problem, we use the ideal gas law to calculate the volume of oxygen gas generated. We take the amount in moles of oxygen gas, the temperature in Kelvin, and the pressure in atmospheres to solve for the volume. To convert the given temperature to Kelvin, we add 273.15 to the Celsius temperature. Similarly, we convert the pressure from torr to atmospheres because the gas constant 'R' uses these units. One important thing to note is that this law assumes all gases behave ideally and deviations can occur under high pressure or at low temperatures. Nevertheless, for educational purposes and under given conditions, its application offers a reasonable approximation. Remember, all the quantities must be in the correct units for the ideal gas law to be applied appropriately.

Molar Mass Determination

Determining the molar mass of a compound is a critical step in stoichiometry and involves the sum of the atomic masses of all atoms in the molecule. For hydrogen peroxide \[(\mathrm{H}_{2}\mathrm{O}_{2})\], this involves adding the molar masses of hydrogen (1.01 g/mol) and oxygen (16.00 g/mol).

Specifically, hydrogen peroxide contains two hydrogen atoms and two oxygen atoms, so its molar mass is calculated as \[2(1.01 \text{ g/mol}) + 2(16.00 \text{ g/mol})\], giving us 34.02 g/mol. When we know the molar mass, we can convert the mass of a substance to moles, a step that is foundational for stoichiometry and imperative for further calculations using the ideal gas law. It's important to be accurate with these calculations, as the molar mass is crucial to converting between grams and moles, dictating the quantities of reactants and products in a given reaction.