Question

A sample of 0.6303 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added in excess. What is the minimum mass of solid and the minimum volume of 2.90 M required to convert all of the ions to ions

Answer 1

For the sample, the minimum mass of solid KI is 2.438 g and the minimum volume of 2.90 M HCl is 6.08 mL.

How to find minimum mass and volume?

Determine the balanced chemical equation:

KIO₃ + 5 KI + 6 HCl → 3 I₂ + 6 KCl + 3 H₂O

Calculate the moles of potassium iodate:

Moles KIO₃ = Mass / Molar mass

= 0.6303 g / 214.01 g/mol

= 0.00294 mol

Calculate the minimum mass of solid KI needed:

From the balanced equation, 5 moles KI react with 1 mole KIO₃.

Therefore, minimum moles KI needed

= 5 × 0.00294 mol

= 0.0147 mol

Minimum mass of KI

= Moles × Molar mass

= 0.0147 mol × 166.01 g/mol

= 2.438 g

Calculate the minimum volume of 2.90 M HCl needed:

From the balanced equation, 6 moles HCl react with 1 mole KIO₃.

Therefore, minimum moles HCl needed

= 6 × 0.00294 mol

= 0.01764 mol

Minimum volume of HCl

= Moles / Molarity

= 0.01764 mol / 2.90 M

= 0.00608 L

= 6.08 mL