Question

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?

Answer 1

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

Answer 2

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 = ?

Answer 3

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.