Question

how many ml of KMNO4 solution containing 158gramper litre must be used to complete the conversion of 75gram of KI to I2 in acidic solution ​

Answer 1

75g / 166.0028 g/mol = 0.4518 moles KI.

From stoichiometric coefficients we know

that 10 moles KI react with 2 moles KMNO4

so 2/10 * 0.4518 moles = 0.09036 moles

KMnO4 are required. Provided with 158 g/L.

158 g/L / 158.054 g/mol = 0.9997 mol/L.

1 L KMnO4 sol'n contains 0.9997mole

so 1000 mL sol'n contains 0.9997 mole.

0.09036 / 0.9997 * 1000 mL = 90.4 mL.