Final answer:
The decomposition of hydrogen peroxide (H2O2) into water and oxygen can be calculated using stoichiometry and the ideal gas law. For 125 g of a 50% by mass hydrogen peroxide solution, 0.92 mol of oxygen gas would be generated. The volume of oxygen gas collected at 27 degrees C and 746 torrs would be approximately 92 mL. The volume of liquid water formed in the reaction would be 1.84 mL.
Step-by-step explanation:
The decomposition of hydrogen peroxide (H2O2) into water and oxygen can be represented by the following equation:
2H2O2 → 2H2O + O2
First, we need to calculate the number of moles of hydrogen peroxide present in the 125 g solution. To do this, we divide the mass by the molar mass of H2O2 (34.01 g/mol) and convert to moles. Mass of H2O2 = 125 g * 0.50 = 62.5 g. Moles of H2O2 = 62.5 g / 34.01 g/mol = 1.84 mol.
According to the stoichiometry of the reaction, 2 moles of H2O2 produce 1 mole of O2. Therefore, 1.84 moles of H2O2 will produce half as many moles of O2. Moles of O2 = 1.84 mol * (1 mol O2 / 2 mol H2O2) = 0.92 mol.
To calculate the volume of oxygen gas at 27 degrees C and 746 torr, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for V (volume), we have V = (nRT) / P. Plugging in the values, V = (0.92 mol * 0.0821 L * atm / K * mol * (27 + 273) K) / 746 torr = 0.092 L or 92 mL.
b. Since water is in liquid form, we need to calculate the volume of liquid water. From the stoichiometry of the reaction, we know that 2 moles of H2O are produced for every 2 moles of H2O2. Therefore, the moles of water produced will be equal to the moles of hydrogen peroxide. Moles of water = 1.84 mol. Now we can convert moles to volume using the density of liquid water (1 g/mL). Volume of water = 1.84 mol * 1 mL/mol = 1.84 mL.