Asked by Santiago
A sample of crude potassium iodide was analyzed using this reaction (not balanced):
I- + SO42- I2 + H2S (acid solution)
If a 4.19-g sample of crude KI produced 1.33 g of iodine, what is the percent purity of the KI?
I- + SO42- I2 + H2S (acid solution)
If a 4.19-g sample of crude KI produced 1.33 g of iodine, what is the percent purity of the KI?
Answers
Answered by
DrBob222
Where's the arrow. I can't tell the products from the reactants.
1. Balance the equation.
2. mols I2 produced = grams/molar mass = ?
3. Using your balanced equation, convert mols I2 to mols I^-
4. Convert mols I^- to mols KI (which will be the same)
5. Convert mols KI to grams KI. g KI = mols KI x molmass KI = ?
6. %purity = (g KI/4.19)*100 = ?
1. Balance the equation.
2. mols I2 produced = grams/molar mass = ?
3. Using your balanced equation, convert mols I2 to mols I^-
4. Convert mols I^- to mols KI (which will be the same)
5. Convert mols KI to grams KI. g KI = mols KI x molmass KI = ?
6. %purity = (g KI/4.19)*100 = ?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.