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

I worked this for you yesterday.

i responded on it but i don't think you saw it

OK. I found the initial post and responded to that. Basically I think you are keying in too many significant figures but I've expanded on that at the initial post.

To determine the number of coulombs of charge required for the electrolysis of water, we need to know the stoichiometry of the reaction and the Faraday's constant.

The balanced equation for the electrolysis of water is:

2 H2O(l) -> 2 H2(g) + O2(g)

From the equation, we can see that for every 2 moles of water electrolyzed, 1 mole of oxygen gas (O2) is produced.

We also know that at STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters of volume.

Therefore, if 10.4 liters of O2 are produced, we can calculate the number of moles of O2 gas as follows:

10.4 liters O2 x (1 mole O2 / 22.4 liters O2) = 0.4643 moles O2

Since the stoichiometry of the reaction is 2 moles of water produces 1 mole of O2, we can infer that 2 x 0.4643 = 0.9286 moles of water were electrolyzed.

Now, to determine the number of coulombs of charge required, we need to use Faraday's constant. Faraday's constant (F) is the charge carried by one mole of electrons, and its value is approximately 96,485 Coulombs per mole of electrons.

Therefore, the number of coulombs of charge required can be calculated as follows:

0.9286 moles H2O x (2 moles of electrons / 1 mole of H2O) x (96,485 Coulombs / 1 mole of electrons) = 178,323.42 Coulombs

So, approximately 178,323.42 Coulombs of charge were required for the electrolysis of 10.4 liters of O2 at STP.