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

                  337 g of Solid PbI₂

                   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

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consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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