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 $Pb(NO_3)_2$.

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}$

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

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 $0.662\times 2=1.32$ moles of $NH_4I$

Now we have to calculate the volume of $NH_4I$

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

$\text{Volume of }NH_4I=\frac{1.32mol}{0.450mol/L}=2.93L$

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.