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 \(27 \mathrm{C}\) and 764 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution?

Short answer

The volume of pure \(O_2\) gas generated by the decomposition of 125 g of a 50% hydrogen peroxide solution is approximately 22.469 L under the given conditions of 27°C and 764 torr.

Step-by-step solution

Calculate moles of hydrogen peroxide in the solution

Given that we have a 50.0% hydrogen peroxide solution, we can find the mass of hydrogen peroxide in the 125 g solution. mass of hydrogen peroxide = 0.50 * 125 g = 62.5 g Now, we need to calculate the moles of hydrogen peroxide. The molar mass of hydrogen peroxide (\(\mathrm{H}_{2} \mathrm{O}_{2}\)) is 34.0147 g/mol (from the periodic table: 2 * 1.00784 - hydrogen + 2 * 15.999 - oxygen). moles of hydrogen peroxide = 62.5 g / 34.0147 g/mol = 1.837 mol

Calculate moles of generated oxygen

Using the balanced chemical equation, we can find the moles of \(O_2\) gas produced. 2 moles \(\mathrm{H}_{2} \mathrm{O}_{2}\) → 1 mole \(O_2\) or 1 mole \(\mathrm{H}_{2} \mathrm{O}_{2}\) → 0.5 mole \(O_2\) So, 1.837 moles of hydrogen peroxide will produce 0.5 * 1.837 moles of \(O_2\), which is equal to 0.9185 moles of \(O_2\).

Calculate the volume of generated oxygen gas

Now we will use the gas law equation (PV = nRT) to find the volume of the generated oxygen gas. Given the temperature T = 27°C (convert to Kelvin: 27 + 273.15 = 300.15 K) and the pressure P = 764 torr (convert to atm: 764 * 1 atm / 760 torr = 1.00526 atm). We know that n = number of moles of \(O_2\) = 0.9185 moles, R (gas constant) = 0.0821 L atm/mol K, T = 300.15 K, P = 1.00526 atm. We need to find the volume V: V = nRT / P V = (0.9185 mol) * (0.0821 L atm/mol K) * (300.15 K) / (1.00526 atm) V ≈ 22.469 L Hence, the volume of pure \(O_2\) gas generated by the decomposition of 125 g of a 50% hydrogen peroxide solution is approximately 22.469 L under the given conditions.

Key concepts

Stoichiometry of Hydrogen Peroxide Decomposition

Stoichiometry is the method of quantifying or measuring elements and compounds involved in chemical reactions. In the decomposition of hydrogen peroxide (H_2O_2), stoichiometry enables us to understand the quantitative relationship between reactants and products.
For every two moles of hydrogen peroxide that decompose, two moles of water (H_2O) and one mole of oxygen gas (O_2) are produced according to the balanced chemical equation:
2 H_2O_2(aq) → 2 H_2O(l) + O_2(g)
This pivotal ratio allows us to predict the amount of oxygen gas produced from a given amount of hydrogen peroxide.
When solving this type of problem, it's important to start by calculating the number of moles of hydrogen peroxide you have. This is done by dividing the mass of H_2O_2 by its molar mass. Next, apply the stoichiometric ratio from the balanced equation to determine the moles of oxygen gas formed. Stoichiometry problems require a stepwise approach to solve, ensuring each conversion is carefully accounted for.

Practical Tips for Stoichiometry:

• Always confirm the chemical equation is balanced before starting your calculations.
• Write out each step to prevent mistakes with ratios.
• Consider units carefully to ensure they cancel appropriately, leaving you with the desired unit of measure.

Applying Gas Laws to Chemical Reactions

Gas laws are mathematical relationships between the pressure (P), volume (V), temperature (T), and quantity in moles (n) of gas. These laws are helpful in predicting how gases will behave under different conditions.
The equation PV = nRT, known as the ideal gas law, consolidates these variables into one formula, where R is the universal gas constant. It works under the assumption that gases behave ideally, meaning their particles have no volume and experience no intermolecular forces.
By rearranging the ideal gas law (as seen in our textbook's solution), we can calculate the volume of oxygen gas produced during the decomposition of hydrogen peroxide, provided we know the other variables:
V = (nRT) / P
To apply this law correctly:

Key Steps:

• Convert all units to the standard ones for gas law equations, i.e., pressure in atm, volume in liters, temperature in Kelvin.
• Calculate the number of moles (n) of gas produced from the reaction, using stoichiometry.
• Ensure the temperature is in Kelvin by adding 273.15 to the Celsius temperature.
• Insert the values into the ideal gas law and solve for the desired variable.

It's critical to grasp how gas properties adjust when conditions change. For instance, as temperature increases, the volume of gas at constant pressure will also expand, epitomizing Charles's Law, which is integral within the ideal gas law's framework.

Molar Mass Calculation in Chemical Reactions

The molar mass of a substance is the mass in grams of one mole of that substance. It's a bridge between the mass of a material and the amount of substance (in moles) and is crucial for stoichiometric calculations in chemistry.
To find the molar mass of hydrogen peroxide (H_2O_2), we add the molar masses of the constituent atoms, measured in grams per mole. Each element's molar mass can be found on the periodic table: hydrogen (H) has a molar mass of approximately 1.00784 g/mol, and oxygen (O) has a molar mass of approximately 15.999 g/mol.
The molar mass calculation for hydrogen peroxide:
2(1.00784 g/mol) + 2(15.999 g/mol) = 34.0147 g/mol

Considerations When Calculating Molar Mass:

• It is vital to use the correct atomic mass for each element from the periodic table.
• Round atomic masses to a reasonable number of decimal places to maintain precision without overcomplicating the calculation.
• Factor in the number of atoms of each element in a molecule to get the total molar mass.

Understanding how to calculate molar mass is key in converting between grams and moles, which is necessary for many stoichiometric computations. This enables precise predictions about the amounts of substances involved in chemical reactions.