Question

PLEASE HELP! I ALREADY TRIED IT AND GOT IT WRONG. An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container.
This is the equation:
CO(g)+H2O(g)--><-- CO2(g) + H2(g)
How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answer 1

I calculated Kc=9

Answer 2

But I have no idea what to do. I would normally write an ICE table but I got it wrong. This is what I did:
CO + H2O <=> CO2 + H2
I 0.2 0.2 0.6 0.6
C +0.1 +0.1 +x -0.1
E 0.3 0.3 0.6+x 0.5

solving for x I got 1.02
And I accidentally put the answer 1.62 and I got it incorrect and it told me that that is the final concentration of CO2. Therefore, I assume I'm on the right path, I just don't know what to do next. PLEASE HELP!!

Answer 3

why did the number of moles of H2 get subtracted???

Answer 4

because, since I'm adding more CO2, the rection is going to shift to the right, so H2 will be subtracted

Answer 5

isnt the answer "1.02"?

Answer 6

you add "1.02" more

Answer 7

I did put that answer and got it wrong

Answer 8

try the answer 1.12moles
I think we missed the effect of increase of CO trying to balance the CO2

Answer 9

no, less would be needed!!
1.02 - 0.1 = 0.92moles.

YES this is it. do it. :)

Answer 10

nope. got it wrong

Answer 11

This is what they show me when I get it wrong

Answer 12

:(
hmm

Answer 13

it was 1.12 -_- hahaha

Answer 14

whaat1!
lol