If 10.4 liters of O2 at STP are produced by the electrolysis of water, how many coulombs of charge were required?

96,485 coulombs will produce 32/2 = 16 g O2 or 11.2L O2 @ STP. Therefore,

C = 96,485 x 10.4/11.2 = ?

i got 89593.2 but its wrong

I suspect you are keying in too many significant figures. 10.4L limits you to 3 s.f.; you're keying in 6 s.f. You may also want to try using 96,500 for 1 Faraday to make that 3 s.f. also.

To calculate the number of coulombs of charge required for the electrolysis of water, we need to use Faraday's law of electrolysis. This law states that the quantity of a substance produced or consumed during an electrolysis reaction is directly proportional to the amount of electric charge passed through the electrolytic cell.

The equation to calculate the amount of charge required is given by:

Q = n * F

Where:
Q is the amount of charge (in coulombs),
n is the number of moles of electrons,
F is Faraday's constant (96485 C/mol).

In the electrolysis of water, the overall reaction is 2 H2O(l) → 2 H2(g) + O2(g). We can see that for every molecule of O2 produced, 4 moles of electrons are required.

First, we need to calculate the number of moles of O2 produced. Since the volume of O2 is given in liters at STP, we can use the ideal gas law to convert it to moles:

n = PV / RT

Where:
P is the pressure (which is 1 atm at STP),
V is the volume of O2 (10.4 L),
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature (which is 273 K at STP).

Substituting the values into the equation, we get:

n = (1 atm)(10.4 L) / (0.0821 L·atm/(mol·K))(273 K)

After performing the calculations, we find that n is approximately 0.478 moles.

Now, we can determine the amount of charge required using Faraday's law:

Q = (0.478 mol * 4) * (96485 C/mol)

Multiplying the number of moles of electrons (0.478 mol) by 4 (since 4 moles of electrons are required for every mole of O2 produced) and then multiplying by Faraday's constant (96485 C/mol), we find that the amount of charge required is approximately 184,067 coulombs.

Therefore, approximately 184,067 coulombs of charge were required to produce 10.4 liters of O2 at STP by the electrolysis of water.