Chemistry Help... Balancing Equations


8 Years
Apr 12, 2011
I am trying to to balance the following reaction by oxidation number method

NaI + H2SO4 -----> Na2SO4 + I2 + H2S +H2O

I split it into two half reactions...

I multiplied the NaI by 2 bc there were two I on the other side; they gave up 2e-

the H2SO4 and H2S gained 5e-

I multiplied by 5 and 2 respectively in order to balance the electrons.

I can't get the equation to balance though! Everything works out except I can't get the H and O to balance! Thanks for any help!
im no chem master, but isnt it NaI + H2SO4 -----> Na2SO4 + HI.
i mean, if its a double displacement reaction, the elements/compounds would just exchange partners, right?
if i'm right, u could balance it by putting a 2 before NaI and a 2 before HI.......again, im no master at this
Well, the equation was given, so I guess it is not a double replacement reaction... Thank you for trying to help. I think I have been working on these darn problems for too long. Maybe in the morning I will look at it and be like, "oh, yeah! duh, that's how you do it..." LOL
i see, thats how it was given....well, its got me stumped. sometimes i feel the same way after ive been working on the same type of problems for a to give the brain a rest, lol. Plus its friday...
2NaI + 2 H2SO4 ----> Na2SO4 + H2S + 2 H2O

It's been a very, very long time, but I think this works out. You get 2 sodium, 2 iodine, 2 hydrogen, 2 sulfur and 8 oxygen on both sides.

If this is right, please let me know. It would make me happy if I could remember how to do this stuff.
Last edited:
The equation above doesn'st balance. It's

8 NaI + 5 H2SO4 = H2S + 4 I2 + 4 Na2SO4 + 4 H2O

This gives 8 sodiums, 8 iodines, 10 hydrogens, 5 sulfurs, and 20 oxygens on both sides.

New posts New threads Active threads

Top Bottom