Question

Calculate the mass of the solid produced when starting with a solution containing 242.36 g of potassium iodide assuming that the reaction goes to completion. Give your answer to three significant figures

Answer 1

337 g of Solid PbI₂

Solution:

When Potassium Iodide (KI) is reacted with Lead (II) Nitrate it produces a yellow precipitates of Lead (II) Iodide (PbI₂) as expressed in following balanced equation.

Pb(NO₃)₂ + 2 KI → K₂(NO₃)₂ + PbI₂

According to this equation,

332 g (2 mol) of KI reacts to produce = 461 g (1 mol) of solid PbI₂

So,

242.36 g of KI will produce = X g of solid PbI₂

Solving for X,

X = (242.36 g × 461 g) ÷ 332 g

X = 336.53 g of Solid PbI₂

Rounded to three significant figures,

X = 337 g of Solid PbI₂