Chemistry: Calculation of Lead Iodide Mass from Potassium Iodide

What is the mass of insoluble lead(II) iodide (461.0 g/mol) produced from 0.830 g of potassium iodide (166.00 g/mol) and aqueous lead(II) nitrate?

A) 4.61 g. B) 0.149 g. C) 2.31 g. D) 1.15 g. E) 0.598 g.

The mass of insoluble lead iodide produced from 0.830 g of potassium iodide in a reaction with lead nitrate is 2.31 g. Thus option C is correct.

Lead iodide or PbI is an ionic compound formed from the ionic bonding between metal lead and iodine. It is industrially prepared from the reaction of aqueous lead nitrate with potassium iodide.

Potassium iodide easily reacts with lead iodide as per the reaction written below:

KI + PbNO_{3} ⟶ KNO_{3} + PbI

As per this reaction one mole of potassium iodide produce one mole of lead iodide. The molar mass of potassium iodide is 166 g/mol and that of lead iodide is 461 g/mol.

Thus, 166 g of potassium iodide gives 461 g of lead iodide. The mass of lead iodide then produced from 0.830 g of potassium iodide is calculated as follows:

mass = (0.830 g × 461)/ 166

= 2.31 g.

Hence, the mass of insoluble lead iodide produced from 0.830 g of potassium iodide in a reaction with lead nitrate is 2.31 g and option C is correct.

← The temperature change of ethanol in a heating process The mystery of oxygen volume a chemical puzzle unveiled →

More Posts: