When hydrogen peroxide decomposes, oxygen is produced: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}(g) $$ What volume of oxygen gas at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) is produced from the decomposition of \(25.00 \mathrm{~mL}\) of a \(30.0 \%\) (by mass) solution of hydrogen peroxide \((d=1.05 \mathrm{~g} / \mathrm{mL}) ?\)

To find the mass of hydrogen peroxide in the solution, we will first determine the mass of the solution and then use the percent by mass to find the mass of hydrogen peroxide. Mass of solution = Volume × Density Mass of solution = 25.00 mL × 1.05 g/mL = 26.25 g Now, we will find the mass of hydrogen peroxide using the percent by mass. Mass of hydrogen peroxide = (Mass of solution × percent by mass) / 100 Mass of hydrogen peroxide = (26.25 g × 30.0%) / 100 = 7.875 g

We will now convert the mass of hydrogen peroxide to moles using its molar mass. Molar mass of hydrogen peroxide (H2O2) = 2 × (1 g/mol) + 2 × (16 g/mol) = 34 g/mol Moles of hydrogen peroxide = Mass / Molar mass Moles of hydrogen peroxide = 7.875 g / 34 g/mol = 0.2316 mol

We will use the balanced chemical equation to find the moles of oxygen produced from the given moles of hydrogen peroxide. 2 H2O2 (aq) → 2 H2O + O2 (g) From the balanced equation, we can see that 2 moles of hydrogen peroxide produce 1 mole of oxygen. Therefore, we can use the stoichiometry to find the moles of oxygen produced. Moles of oxygen = (Moles of hydrogen peroxide × moles of oxygen produced) / moles of hydrogen peroxide consumed Moles of oxygen = (0.2316 mol × 1 mol O2) / 2 mol H2O2 = 0.1158 mol

Finally, we will use the ideal gas law to find the volume of oxygen gas produced. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We are given the temperature in Celsius, so we need to convert it to Kelvin. Temperature in Kelvin = Temperature in Celsius + 273.15 Temperature in Kelvin = 25°C + 273.15 = 298.15 K We will use the ideal gas law to find the volume (V) of the oxygen gas produced, given that the pressure (P) is 1.00 atm, the number of moles (n) is 0.1158 mol, the gas constant (R) is 0.08206 L atm/mol K, and the temperature (T) is 298.15 K. V = nRT / P V = (0.1158 mol × 0.08206 L atm/mol K × 298.15 K) / 1.00 atm V = 2.800 L The volume of oxygen gas produced at 25°C and 1.00 atm is approximately 2.800 L.