Help needed

[Drilon]

I'v had this problem.

 

How much iodine will we make from 350gr Sodium iodate reacting with sodium hydrogen sulphide.

 

Sorry about my english, hope I wrote it good.

[chem1a]

Calculations for Iodine available not the iodine given from this reaction .... i dont know how these chems react sorry ... if u give an eqn i can calculate closer to actual yield

 

 

NaIO3

 

Iodate Anion

Iodine = 126.9g per mole

Oxygen=15.9 per mole & we have 3 parts oxygen in this compound * 3 = 47.7

therefore

174.6 G for 1 mole

 

Sodium Cation

Na 22.99g per mole

 

 

1 mole substance = 197.59

 

BUT you have HAVE 350 g

so 350g/197.59g = moles of substance

= 1.77moles

Mol Iodine relationship to Mol NaIO3 is 1:1

therefore max theoretical amount of iodine available 1.77mol * 126.9g = 224.8g Iodine present

 

Also in that sample is Na 40.7g

and 84.5G Oxygen

 

(Test the math) 84.5 + 40.7 + 224.8 = 350 (100% of the original compositon)