when electricity is passed through water, the water decomposes into hydrogen gas and oxygen gas. when 10.00 kg of water were decomposed this way, 5788 l of oxygen gas were produced at 25 degrees celsius and 1 atm pressure. what is the percent yield of oxygen by volume?
H2O = 18 g/mol, 16 of which is O
so we have in the water (16/18)(10) = 8.89 kg of O
that is 8890 grams or 556 mols of O
or 278 mols of O2 in the water
how many mols of O2 did we end up with in the air
T = 273+25 = 298
do P V = n r T, solve for n
or know that 1 mol is 22.4 liters at STP
we are close to STP so 5788/22.4 = 258 mols
percent = 100 (258/278) = 93 % yield
You need to calculate the theoretical yield.
2H2O ==> 2H2 + O2
mols H2O = grams/molar mass = 10,000/18 = ?
Using the coefficients in the balanced equation, convert mols H2O to mols O2. That's ? mols H2O x (1 mol O2/2 mols H2O) = ? mols H2O x 1/2 = ?
Then convert mols O2 to L O2. L = mols x 22.4 L/mol. This is the theoretical yield (TY). The actual yield (AY) is given in the problem as 5788 L.
% yield = (AY/TY)*100 = ?
I used 22.4 L/mol for O2 and that is true for 273 C but for 298 I should have used 24.45 L/mol. That is, mols O2 x 24.45 L/mol = ? L O2 and that's the theoretical yield.
To calculate the percent yield of oxygen gas by volume, you need to compare the actual volume of oxygen gas produced with the theoretical volume of oxygen gas that should have been produced based on stoichiometry.
Let's start by calculating the theoretical volume of oxygen gas using the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (1 atm)
V = volume in liters (unknown)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L*atm/(mol*K))
T = temperature in Kelvin (25 + 273 = 298 K)
Rearranging the equation to solve for volume (V), we have:
V = (nRT) / P
Next, we need to determine the number of moles of oxygen gas produced. Since we know the volume of oxygen gas (5788 L) and the conditions (25 degrees Celsius), we can apply the ideal gas law again to find the number of moles:
PV = nRT
Substituting the given values:
(1 atm) * (5788 L) = n * (0.0821 L*atm/(mol*K)) * (298 K)
Solving for n:
n = (1 atm * 5788 L) / (0.0821 L*atm/(mol*K) * 298 K)
Now that we have the number of moles (n) of oxygen gas produced, we can calculate the theoretical volume (V) of oxygen gas:
V = (nRT) / P
V = [(1 atm * 5788 L) / (0.0821 L*atm/(mol*K) * 298 K)] * (0.0821 L*atm/(mol*K) * 298 K) / (1 atm)
V = 5788 L
The theoretical volume of oxygen gas produced is 5788 L.
Now we can calculate the percent yield of oxygen gas by volume:
Percent Yield = (Actual volume / Theoretical volume) * 100
Percent Yield = (5788 L / 5788 L) * 100
Percent Yield = 100%
The percent yield of oxygen gas by volume is 100%.