[bmcevoy1776]

A student places a .400 mol sample of liquid CH3OH
into an evacuated, rigid 1liter
container and heats the container to 550 kelvin
. At this temperature, the methanol is completely vaporized and begins to decompose according to the equation above. When equilibrium is reached, 60.5 
 percent of the original number of moles of CH3OH
 have decomposed.
Using molar concentrations, calculate the value of the equilibrium constant, Kc , for the reaction at 550 kelvin
 Write your answer using three significant figures.
my problem is nothing explicitly tells us which 1 of the 2 products are H2. yes 60% of .4 mol is .242. & yes - from .4 is  .158. but no context clues tell you which? the answer could just as easily be .1793 if .242 was squared. what context clues does it give? even chatGPT cant calrify.

[Hunter2]

First you need the chemical equation and the formula of the  law of mass action.
Kc =..
In equilibrium you have to fill in the remaining CH3OH and the new amount of the both products.

[Borek]

decompose according to the equation above

You haven't listed the equation, so it is difficult to help, but in general - follow the stoichiometry. You know how many moles decomposed, you should be easily able to calculate numbers of moles of products.

even chatGPT cant calrify.

Even don't try realying on ChatGPT in such cases. It is completely useless for this kind of problems, and even when it states something it is often completely wrong. Compare https://www.chemicalforums.com/index.php?topic=113085.0

[bmcevoy1776]

CH3OH -> CO2 + 2H2. sorry. my point was the stoichiometric coefficient of H2 is 2. but why is it assigned to 1 (.242) & not the other(.484). i use chatgpt because numerous chemistry outlets dont explain the theorem they used. so if your not an absolute ace at algebra then they don't elaborate.
what is the context or que to assign H2's coefficient to.484 rather than .242??

[Hunter2]

You have CH3OH => CO + 2 H2

You have a ratio 1 to 1 to 2

Instead of 1 mol you have only 60.5% of 0.4 mol what is 0.242 mol

So how much CO and H2 you have?

[Borek]

i use chatgpt because numerous chemistry outlets dont explain the theorem they used.

ChatGPT doesn't either - it just pretends to, and if you don't know the subject you'll never know when it starts to hallucinate. Really, really bad idea.

And - sorry to say that - if you want to be reasonably good at chemistry, you need to be reasonably good at math. No way around.

60.5% of 0.4 moles reacted, that's 0.242 moles - so far, so good.

Then it is all in a trivial stoichiometry. Do you know what stoichiometric coefficients mean, and how to read the reaction equation?

Compare https://www.chembuddy.com/balancing-stoichiometry-stoichiometric-calculations

[bmcevoy1776]

being condescending isn't an explanation. chatgpt has helped greatly to clarify the theorems being used.  that's not the point. what context is given to assign the 2H2 to either .242 or .484? that's the question. if you don't know then don't comment. 

[Hunter2]

Sorry you have no idea of Chemistry I think
From 1 Methanol you get 1 carbonoxide and 2 hydrogen.
So if you have 0.242 mol CH3OH you get also the same 0.242 mol CO and the double H2 what means 0.484 mol. That is normal stoichometry.
With this hint now try to get the constant Kc.

[bmcevoy1776]

Khan academy still shows the stoichiometric coefficient for the final equation which led to the incorrect answer.
https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:equilibrium/x2eef969c74e0d802:calculating-equilibrium-concentrations/e/calculating-equilibrium-concentrations 
in this example .436 raised to the 1/2 power is .218. but they show .218 raised to the 1/2 power again in the final theorem . not my fault they explain it like assholes. this would be raising it to the stoichiometric coefficient 2 different times.

[bmcevoy1776]

here is the other problem with a like example of raising to the coefficient twice for no reason that is elaborated upon 

[Hunter2]

Your link guide to a exercise of Hydrogeniodide.

The second one is based of

NOBr => NO + 0.5 Br2
1 mol Nitrosylbromide decompose to 1 mol nitrogenmonoxid and a half mol of bromine.
So 54,5 % is 0.436 from 0.8 mol and this is equal to the amount of NO and the half of Bromine.
Where do you see a problem.
It is similar exercise like the example of the Methanol.

[bmcevoy1776]

ss

[bmcevoy1776]

its raises to the stochiometric coefficient twice. were am i supposed to derive that from? i get if there are 2 of them raise to the second power. but why am i doing it a second time?

[Hunter2]

I dont know what you mean:

its raises to the stochiometric coefficient twice

Where is it rised two times.

[bmcevoy1776]

.436 is what is decomposed so its reasonable to assume that is the product. 436 raised to the 1/2 power is how you get .218.
but to get the answer you multiply .436*.218^2 & divided by .364 its raised to the second twice with no explanation.  its in the picture