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lead iodide is virtually insoluble in water so that the reaction appears to go to completion. how many milliliters of 3.550 m hi(aq) must be added to a solution containing 0.700 mol of pb(no3)2 (aq) to completely precipitate the lead?

Respuesta :

The volume of HI required to react with Pb(NO3)2 to completely precipitate the Lead is 0.197 ml.

First, let us write the balanced chemical equation of the reaction of HI with Pb(NO3)2,

HI + Pb(NO3)2 = PbI2 + HNO3

From the reaction, we can see,

One mole of HI reacts with one mole of Pb(NO3)2 to completely precipitate the Less Iodide solution.

We also know,

Mole = molarity × volume of solution.

So,

Moles of Pb(NO3)2 are 0.700 moles.

Let us say the volume 3.55M of HI used is V,

So,

We can write,

0.700 = 3.55 × Volume

Volume = 0.197 ml.

So, 0.197 ml of HI is required to completely precipitate the lead solution.

To know more about molarity, visit,

https://brainly.com/question/14469428

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