You have 25.00 mL of 3% by mass solution of hydrogen peroxide. Assume the density of this solution is 1.00 g/mL. If all of the hydrogen peroxide decomposes to water and oxygen according to the reaction below:
2H2O2 (aq) <--> 2H2O (l) + O2 (g)
What is the total volume of oxygen gas generated if the temperature is 24 degrees C and the pressure is 770 Torr. (R = 0.08206 L atm/K mol) At this point you may ignore the effect of water vapor.
To find the total volume of oxygen gas generated, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.08206 L atm/K mol)
T = temperature (in Kelvin)
First, let's convert the given pressure from Torr to atm:
770 Torr = 770/760 atm ≈ 1.01316 atm
Next, we need to calculate the number of moles of oxygen gas produced. From the balanced equation, we can see that for every 2 moles of hydrogen peroxide, 1 mole of oxygen gas is produced.
Given that the volume of the hydrogen peroxide solution is 25.00 mL and its density is 1.00 g/mL, we can determine the mass of the solution:
mass = volume × density = 25.00 mL × 1.00 g/mL = 25.00 g
Next, we need to calculate the mass of the hydrogen peroxide in the solution. Since the solution is 3% by mass hydrogen peroxide, we can multiply the mass of the solution by 0.03:
mass of hydrogen peroxide = 25.00 g × 0.03 = 0.75 g
Now, we can calculate the number of moles of hydrogen peroxide using its molar mass:
molar mass of H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles of H2O2 = mass of H2O2 / molar mass of H2O2 = 0.75 g / 34.02 g/mol ≈ 0.022 moles
Since 2 moles of hydrogen peroxide produce 1 mole of oxygen gas, we can find the number of moles of oxygen gas produced:
moles of O2 = moles of H2O2 / 2 = 0.022 moles / 2 ≈ 0.011 moles
Now, we have all the necessary values to calculate the volume of oxygen gas:
P = 1.01316 atm
V = volume of oxygen gas (unknown)
n = 0.011 moles
R = 0.08206 L atm/K mol
T = 24 degrees C = 24 + 273.15 K = 297.15 K
Rearranging the ideal gas law equation, we get:
V = nRT / P
Substituting the values:
V = (0.011 moles) × (0.08206 L atm/K mol) × (297.15 K) / (1.01316 atm)
V ≈ 0.2478 L
Therefore, the total volume of oxygen gas generated is approximately 0.2478 liters.