[SunnyBlu]

Hello, I've been having a lot of trouble with this problem and would really appreciate some help solving it:
If 13.5 mL of 58% HSO−3   by mass is obtained in the reaction, what mass of S2O2− was present
initially, assuming the reaction went to completion? The density of the HSO−3 solution is 1.45g/cm3.
The original reaction is:
2 S2O2−(aq) + 2 H3O+(aq) →
2HSO−3(aq)+2H2O(l)+2S(s)
 It's apparently a disproportionation redox reation
I thought all you had to do was get the grams of HSO3- and then the moles and since the ration is 1:1 multiply the moles by the mass of S2O2, but that was wrong.
And then I thought of getting the mass of HSO3- by dividing by 58% or .58 and getting the grams that way
but that was wrong too

I would really appreciate any advice
Thanks in advance!

[opti384]

Show what you did. You have density and volume, therefore you can find the mass of the HSO−3  solution and from the percent concentration find the mols.

[Borek]

S₂O^2- or S₂O₂^-? As far as I know, neither exists.

What is HSO-3? Perhaps HSO₃^-?

Even if so, density and concentration given are not those of sulfurous acid solution, but those of sulfuric acid solution. Sulfurous acid is unstable and can't be concentrated to 58%.

Complete chaos.

[SunnyBlu]

Sorry, my copy&paste is obnoxious
The reaction is 2S2O3 with a -2 charge(aq) + 2H3O+(aq)  2HSO3 with a -1 charge(aq) + 2H2O(aq) + 2S(s)

The other numbers are correct. I think my prof gives us unrealistic numbers to work with so that we can't cheat off the internet or something.

But anyway, so I got the moles of HSO3- by multiplying 13.5 times 1.45 to get grams, because 1 mL = 1cm3. And I got 19.575
SO I thought that by saying 58% percent was yielded, that we had to find the whole. So i divided 19.575/.58 and got 33.75 and then divided by the molar mass of HSO3, 81.11, so .416 moles. Then i multiplied by the molar mass of S2O3 to get 46.68 grams. And left it like that since the ratio is 2:2 or 1:1.

D: Sorry about the confusion and thanks for helping!!

[Borek]

But anyway, so I got the moles of HSO3- by multiplying 13.5 times 1.45 to get grams, because 1 mL = 1cm3. And I got 19.575
SO I thought that by saying 58% percent was yielded, that we had to find the whole.

What is the definition of percent concentration?

The way you approached it you have mass of substance dissolved higher than the mass of the solution. Does it make sense?