Question

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb ( NO 3 ) 2 ( aq ) + 2 NH 4 I ( aq ) ⟶ PbI 2 ( s ) + 2 NH 4 NO 3 ( aq ) What volume of a 0.450 M NH4I solution is required to react with 945 mL of a 0.700 M Pb(NO3)2 solution? volume: mL How many moles of PbI2 are formed from this reaction?

Answer 1

Answer : The volume of
`NH_4I` solution required is, 2.93 L

The number of moles of
`PbI_2` formed from the reaction is, 0.662 moles.

Explanation :

First we have to calculate the initial moles of
`Pb(NO_3)_2`.

`\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2* \text{Volume of solution}`

`\text{Moles of }Pb(NO_3)_2=0.700M* 0.945L=0.662mol`

Now we have to calculate the moles of
`NH_4I`

The balanced chemical reaction is:

`Pb(NO_3)_2(aq)+2NH_4I(aq)\rightarrow PbI_2(s)+2NH_4NO_3(aq)`

From the balanced chemical reaction we conclude that,

As, 1 mole of
`Pb(NO_3)_2` react with 2 moles of
`NH_4I`

So, 0.662 mole of
`Pb(NO_3)_2` react with
`0.662* 2=1.32` moles of
`NH_4I`

Now we have to calculate the volume of
`NH_4I`

`\text{Volume of }NH_4I=\frac{\text{Moles of }NH_4I}{\text{Concentration of }NH_4I}`

`\text{Volume of }NH_4I=(1.32mol)/(0.450mol/L)=2.93L`

Now we have to calculate the moles of
`PbI_2`

From the balanced chemical reaction we conclude that,

As, 1 mole of
`Pb(NO_3)_2` react to give 1 moles of
`PbI_2`

So, 0.662 mole of
`Pb(NO_3)_2` react to give 0.662 moles of
`PbI_2`

Thus, the number of moles of
`PbI_2` formed from the reaction is, 0.662 moles.