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* \text{Volume of solution}=0.640M* 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* 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}=(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