[Johan]

I didn't want to take up more forum space, but there's another problem i'm having issues with again.

A soluble Iodide was dissolved in water. Then an excess of silver(I) nitrate, AgNO3, was added to completely precipitate out all of the iodine ion as silver(I) iodide, AgI. If 1.545 grams of soluble iodide gave 2.185 of silver(I) iodine, then

(a) How many moles of I(I-) are in the original soluble iodide sample?
(b) How many grams of I(I-) are in the original soluble sample?
(c) What is the percentage mass of I in the original soluble iodine sample?

Wouldn't 1.545g of soluble iodide all come out on the other side?
When i do the math 126.9g I/234.8g AgI = 54.05%

                            2.185 X .5405 = 1.181g I  , so the way i see this.. this is all of the Iodide on the reactant side... but that doesn't equal 1.545g.. what else is with the soluble iodide? The part that says 1.545g of soluble iodide is throwing me off.. it appears to me as if they gave me the answer and its not working out.. can anyone clarify? Or am i supposed to assume that 1.545g is Water and Soluble iodide together?

[Borek]

You need some cation in iodide, don't you? Like potassium, or something

[Johan]

So do i just ignore this and assume there is a cation and i don't need to know it? I mean.. i guess i don't really need to know it.. but it bothers me.

[LQ43]

(c) What is the percentage mass of I in the original soluble iodine sample?


                            2.185 X .5405 = 1.181g I  , so the way i see this.. this is all of the Iodide on the reactant side... but that doesn't equal 1.545g.. what else is with the soluble iodide? The part that says 1.545g of soluble iodide

you're almost there, yes you can ignore the cation, no need to know what it is, just know that its there.

you have all the info to calculate the answer for c)