Hydrogen peroxide, H2O2, is used to disinfect contact lenses. What volume of O2, in mL, or O2(g) at 31 degree celsius, and 741mmHg can be liberated from 10.0mL of an aqueous solution containing 3.00% H2O2 by mass?
The density of aqueous solution of H2O2 is 1.01g/mL.
Thanks! :)
10.0 mL H2O2 x 1.01 g/mL x 0.03 = ? g O2.
mols O2 = g/molar mass
Then use PV = nRT to solve for volume O2 at the conditions listed.
56.7
To find the volume of oxygen gas (O2) that can be liberated from 10.0 mL of the aqueous solution containing 3.00% H2O2 by mass, we need to follow these steps:
Step 1: Calculate the mass of H2O2 in the solution
Given that the aqueous solution of H2O2 has a density of 1.01 g/mL, we can find the mass of H2O2 in 10.0 mL of the solution as follows:
Mass of H2O2 = Volume of solution x Density of solution
Mass of H2O2 = 10.0 mL x 1.01 g/mL
Mass of H2O2 = 10.1 g
Step 2: Convert the mass of H2O2 to moles
To convert the mass of H2O2 to moles, we need to know the molar mass of H2O2, which is 34.0147 g/mol. We can use this information to calculate the moles of H2O2 as follows:
Moles of H2O2 = Mass of H2O2 / Molar mass of H2O2
Moles of H2O2 = 10.1 g / 34.0147 g/mol
Moles of H2O2 ≈ 0.2970 mol
Step 3: Use the mole ratio to determine moles of O2 produced
According to the balanced chemical equation:
2 H2O2 → 2 H2O + O2
From the equation, we can see that 2 moles of H2O2 produce 1 mole of O2. Therefore, the moles of O2 produced can be calculated as follows:
Moles of O2 = (Moles of H2O2) / 2
Moles of O2 = 0.2970 mol / 2
Moles of O2 = 0.1485 mol
Step 4: Convert moles of O2 to volume at STP using the ideal gas law
We need to convert the moles of O2 gas to volume at STP (Standard Temperature and Pressure). The conditions for STP are a temperature of 0 degrees Celsius (273.15 K) and a pressure of 1 atm.
Using the ideal gas law equation (PV = nRT), we can find the volume as follows:
Volume of O2 at STP = (Moles of O2) x (RT/P)
Where:
R is the ideal gas constant = 0.0821 L.atm/(mol.K)
T is the temperature in Kelvin (273.15 K in this case)
P is the pressure in atm (in this case, 1 atm)
Volume of O2 at STP = (0.1485 mol) x (0.0821 L.atm/(mol.K)) x (273.15 K) / (1 atm)
Volume of O2 at STP ≈ 3.725 L
Step 5: Convert volume of O2 at STP to volume at given conditions using the combined gas law
The combined gas law formula is:
(P1V1)/T1 = (P2V2)/T2
We can rearrange this formula to solve for V2 (the volume of O2 at the given conditions):
V2 = (P1V1 x T2) / (P2 x T1)
Given:
T1 = 273.15 K (0 degrees Celsius to Kelvin conversion)
T2 = 31 + 273.15 K (convert 31 degrees Celsius to Kelvin)
P1 = 1 atm (STP pressure)
P2 = 741 mmHg (given pressure)
V1 = 3.725 L (volume of O2 at STP calculated in Step 4)
V2 = (1 atm x 3.725 L x (31 + 273.15) K) / (741 mmHg x 273.15 K)
V2 = 0.1475 L
Step 6: Convert volume of O2 from liters to milliliters
Finally, to convert the volume of O2 from liters to milliliters, we multiply by a conversion factor of 1000 mL/L:
Volume of O2 = 0.1475 L x 1000 mL/L
Volume of O2 = 147.5 mL
Therefore, the volume of O2 that can be liberated from 10.0 mL of the aqueous solution containing 3.00% H2O2 by mass at 31°C and 741 mmHg is approximately 147.5 mL.
To find the volume of O2 gas liberated from the given aqueous solution of H2O2, we need to follow these steps:
1. Determine the mass of H2O2 present in 10.0 mL of the solution.
2. Calculate the moles of H2O2 using its molar mass.
3. Use the balanced chemical equation for the decomposition of H2O2 to calculate the moles of O2 gas produced.
4. Apply the ideal gas law to convert the moles of O2 gas to volume using the given temperature and pressure.
Let's go through each step in detail:
Step 1: Determine the mass of H2O2
Given that the solution is 3.00% H2O2 by mass, we can find the mass of H2O2 in the solution using the density:
Mass of solution = Volume of solution x Density
Mass of solution = 10.0 mL x 1.01 g/mL
Mass of solution = 10.1 g
Mass of H2O2 = 3.00% of Mass of solution
Mass of H2O2 = 3.00% x 10.1 g
Mass of H2O2 = 0.303 g
Step 2: Calculate moles of H2O2
The molar mass of H2O2 (hydrogen peroxide) is:
2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
Moles of H2O2 = Mass of H2O2 / Molar mass
Moles of H2O2 = 0.303 g / 34.02 g/mol
Moles of H2O2 = 0.008925 mol
Step 3: Use the balanced chemical equation
The balanced chemical equation for the decomposition of H2O2 is:
2 H2O2(aq) -> 2 H2O(l) + O2(g)
From the equation, we can see that 2 moles of H2O2 produce 1 mole of O2 gas.
Moles of O2 gas = Moles of H2O2 / 2
Moles of O2 gas = 0.008925 mol / 2
Moles of O2 gas = 0.0044625 mol
Step 4: Apply the ideal gas law
PV = nRT
Where:
P = pressure (741 mmHg)
V = volume (what we need to find)
n = moles of gas (0.0044625 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin (31 °C + 273.15 = 304.15 K)
Rearranging the equation to solve for V:
V = (nRT) / P
V = (0.0044625 mol x 0.0821 L•atm/mol•K x 304.15 K) / 741 mmHg
Converting mmHg to atm:
V = (0.0044625 mol x 0.0821 L•atm/mol•K x 304.15 K) / (741 mmHg x 1 atm/760 mmHg)
Finally, solving for V:
V = 0.0125 L = 12.5 mL
Therefore, the volume of O2 gas liberated from the 10.0 mL aqueous solution containing 3.00% H2O2 by mass is 12.5 mL at 31 °C and 741 mmHg.