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?
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 = ?