Question

Lead(ii) nitrate and ammonium iodide react to form lead(ii) iodide and ammonium nitrate according to the reaction pb(no3)2(aq)+2nh4i(aq)⟶pbi2(s)+2nh4no3(aq) what volume of a 0.610 m nh4i solution is required to react with 229 ml of a 0.640 m pb(no3)2 solution?

Answer 1

  0.229 L* 0.640 mol Pb(No3)2/L  =0.147 mol Pb(No3)2
by equation 1 mol Pb(No3)2 requires 2 mol NH4I
           so     0.147 mol Pb(No3)2  requires 2*0.147 mol NH4I 

2*0.147 mol NH4I = 0.294 mol NH4I
 0.294 mol NH4I*  1L/0.610 mol =0.482 L =482 ml of the 0.610 m NH4i solution

Answer 2

Answer : The volume of $NH_4I$ solution required is 482 mL

Explanation :

The given balanced chemical reaction is:

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

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

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}=0.640M\times 0.229L=0.147mole$

Now we have to calculate the moles of $NH_4I$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react to give 2 moles of $NH_4I$

So, 0.147 mole of $Pb(NO_3)_2$ react to give $2\times 0.147=0.294$ moles of $NH_4I$

Now we have to calculate the volume of $NH_4I$

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

$\text{Volume of solution}=\frac{0.294mole}{0.610mole/L}=0.482L=482mL$

conversion used : (1 L = 1000 mL)

Therefore, the volume of $NH_4I$ solution required is 482 mL