What STP volume of oxygen gas is obtained from the electrolysis of 140.0mLof water? (Hint: The density of water is 1.00g/mL.)
To find the volume of oxygen gas obtained from the electrolysis of water, we need to use the concept of stoichiometry and the ideal gas law at standard temperature and pressure (STP).
First, let's write the balanced equation for the electrolysis of water:
2H₂O(l) → 2H₂(g) + O₂(g)
The equation tells us that for every 2 moles of water, we obtain 1 mole of O₂ gas.
Next, we can calculate the number of moles of water using its volume and density. Given the volume of water is 140.0 mL and the density of water is 1.00 g/mL, we can convert the volume of water to grams:
Mass of water = Volume of water × Density
Mass of water = 140.0 mL × 1.00 g/mL = 140.0 g
Now, we can convert the mass of water to moles using the molar mass of water. The molar mass of water (H₂O) is:
2(1.01 g/mol of H) + 16.00 g/mol of O = 18.02 g/mol
Number of moles of water = Mass of water / Molar mass of water
Number of moles of water = 140.0 g / 18.02 g/mol ≈ 7.772 mol
According to the balanced equation, for every 2 moles of water, we obtain 1 mole of O₂ gas. Therefore, the number of moles of O₂ gas produced will be half of the moles of water.
Number of moles of O₂ gas = 1/2 × Number of moles of water
Number of moles of O₂ gas = 1/2 × 7.772 mol ≈ 3.886 mol
Finally, we can use the ideal gas law at STP to calculate the volume of O₂ gas.
The ideal gas law is given by:
PV = nRT
Where:
P = Pressure (STP: 1 atm)
V = Volume of the gas
n = Number of moles of the gas
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin (STP: 273 K)
Rearranging the ideal gas law to solve for volume (V), we have:
V = nRT / P
Substituting the known values:
V = 3.886 mol × 0.0821 L·atm/mol·K × 273 K / 1 atm
V ≈ 87.5 L
Therefore, the volume of oxygen gas obtained from the electrolysis of 140.0 mL of water is approximately 87.5 L at STP.
2H2O ==> 2H2 + O2
Use density to convert 140 mL H2O to grams.
Convert g H2O to mols. mols = grams/molar mass.
Use the coefficients in the balanced equation to convert mols H2O to mols O2.
Then L O2 = mols O2 x 22.4 L/mol