Question

When hydrogen peroxide decomposes, oxygen is produced: 2H2O2(aq) 9: 2H2O 1 O2(g) What volume of oxygen gas at 258C and 1.00 atm is produced from the decomposition of 25.00 mL of a 30.0% (by mass) solution of hydrogen peroxide (d 5 1.05 g/mL)? 39\. Ammonium nitrate can be used as an effective explosi

Short answer

Answer: Approximately 4.960 liters of oxygen gas is produced.

Step-by-step solution

Calculate the mass of hydrogen peroxide

We are given that the volume of hydrogen peroxide solution is 25.00 mL, the density is 1.05 g/mL and the concentration is 30.0% by mass. To find the mass of hydrogen peroxide in the solution, multiply the volume, density, and mass percentage together: Mass of H2O2 = Volume × Density × Mass percentage = 25.00 mL × 1.05 g/mL × 0.30 = 7.875 g

Find the moles of hydrogen peroxide

We can now convert the mass to moles using the molar mass of H2O2: Moles of H2O2 = Mass / Molar mass The molar mass of hydrogen peroxide is 34.02 g/mol (2 Oxygen atoms 16.g/mol and 2 hydrogen atoms 1.01 g/mol). Moles of H2O2 = 7.875 g / 34.02 g/mol = 0.2314 moles

Find the moles of oxygen gas produced

The balanced equation tells us that 2 moles of H2O2 will produce 1 mole of O2. Using the stoichiometric ratio, we can determine the moles of O2 produced: Moles of O2 = Moles of H2O2 × (1 mole O2 / 2 moles H2O2) = 0.2314 moles × (1/2) = 0.1157 moles

Calculate the volume of oxygen gas produced using the ideal gas law

Now, we will use the ideal gas law (PV=nRT) to find the volume of O2 produced at 258°C and 1.00 atm. First, we need to convert the temperature to Kelvin: T = 258°C + 273.15 = 531.15 K The ideal gas constant, R, is 0.0821 L·atm/mol·K. Using the ideal gas law, we can solve for volume (V): PV = nRT V = nRT / P V = (0.1157 moles) × (0.0821 L·atm/mol·K) × (531.15 K) / (1.00 atm) V ≈ 4.960 L So, the volume of oxygen gas produced is approximately 4.960 liters.

Key concepts

Stoichiometry

Understanding stoichiometry is crucial in solving chemistry problems, especially when we are dealing with chemical reactions and want to know how much of a reactant is needed or how much of a product is formed. In the context of the hydrogen peroxide decomposition exercise, stoichiometry allows us to calculate the exact amount of oxygen gas produced from a given amount of hydrogen peroxide.

Let's recall the balanced chemical equation for this reaction:
2 H₂O₂(aq) → 2 H₂O(l) + O₂(g)

Here, the coefficients indicate the proportion by which the substances react and are produced. The equation shows that two molecules of hydrogen peroxide yields one molecule of oxygen gas. Therefore, we use this stoichiometric relationship to calculate the moles of oxygen generated from the moles of hydrogen peroxide. This step is vital as it sets the foundation for working out the volume of gas produced which is explained further when we apply the ideal gas law.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas with the universal gas constant (R). Expressed as PV = nRT, it's a powerful tool to predict the behavior of gases under various conditions. For our exercise, it allows us to determine the volume of oxygen produced at specific temperature and pressure conditions once we've found the moles of oxygen.

To apply the ideal gas law, we considered the temperature in Kelvin, which we can calculate by adding 273.15 to the Celsius temperature. Using the given conditions (25°C and 1.00 atm), we calculated the volume of the oxygen gas produced. Additionally, careful attention must be paid to the units used when applying the equation to ensure they are consistent with the gas constant's units (L·atm/mol·K).

By reiterating the calculation of the volume of the gas produced, we reinforce the importance of understanding the ideal gas law, not just in this problem but in explaining the behavior of gases in general.

Mole Concept

The mole concept is a foundational principle in chemistry that provides a bridge between the atomic or molecular scale and the macroscopic world. One mole is equivalent to Avogadro's number of particles (atoms, molecules, ions, or electrons), which is approximately 6.022 x 10²³ particles. In the hydrogen peroxide decomposition example, we use the mole concept to convert the mass of hydrogen peroxide to moles by using its molar mass.

The molar mass of hydrogen peroxide (H₂O₂) is the sum of the masses of its constituent atoms: 2(Hydrogen) + 2(Oxygen) = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol.

By dividing the mass of hydrogen peroxide by its molar mass, we obtain the moles, which are necessary for stoichiometric calculations and subsequently applying the ideal gas law to find the desired volume of oxygen gas. The mole concept is at the heart of chemical calculations and understanding it is essential for anyone studying chemistry.