When a sample of I(g) (18.15 mol) is placed in 59.00 L reaction vessel at 605.0 °C and allowed to come to equilibrium the mixture contains 6.938 mol of I2(g). What is the concentration (mol/L) of I(g)?

Question

2I(g) = I2(g)

Answer 1

It appears that all of these posts (this one and the next half dozen or so) are the same kind of problem. Here is how the first one is done. Use this as a template for the others.If you have questions about this one or the others post your work and explain in detail what kind of trouble you are having.

.............2I(g) ==> I2
I...........18.15......0
C............-2x.......x
E.........18.15-2x.....x

The problem tells you x = 6.938 mols I2.
Then mols I = 18.15-2x = ?
M I2 = mols I2/59.00L = ?
M I = mols I/59.00L = ?

Answer 2

My issue is that I took the first semester of general chemistry almost 2 years ago. I have forgotten everything. This is why everything is brand new to me... Can you recommend any online reading to catch up?

So for this problem this is what I get... Please let me know what I did wrong.
Then mols I = 18.15-2x = 4.74
so MI = 4.74/59 = .0724?

Answer 3

So for this problem this is what I get... Please let me know what I did wrong.
Then mols I = 18.15-2x = 4.74
so MI = 4.74/59 = .0724?

18.15-2x is not 4.74. You must have punched the wrong buttons on the calculator.
18.15 - (2*6.938) = 4.274. In subtraction you are allowed to the hundredths place so I would round to 4.27.
Then 4.27/59.00 = 0.0724 M. That's what you have; therefore, I suspect you just copied the 4.74 wrong or it's a typo.

Answer 4

thank you dr. bob!