Question

How many ml of 0.112 M Pb(NO3)2 are needed to completely react with 10.0ml of a 0.105 M KI Given Pb(NO3)2 (aq) + 2KI(aq)=2KNO3(aq)?

Answer 1

Answer : The volume of $Pb(NO_3)_2$ needed are, 4.6875 ml

Explanation :

First we have to calculate the moles of $KI$.

$\text{Moles of }KI=\text{Molarity of }KI\times \text{Volume of solution}=0.105mole/L\times 0.01L=0.00105mole$

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

The given balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+PbI_2$

From the balanced chemical reaction, we conclude that

As, 2 moles of KI react with 1 mole of $Pb(NO_3)_2$

So, 0.00105 moles of KI react with $\frac{0.00105}{2}=0.000525$ mole of $Pb(NO_3)_2$

Now we have to calculate the volume of $Pb(NO_3)_2$

$\text{Volume of }Pb(NO_3)_2=\frac{\text{Moles of }Pb(NO_3)_2}{\text{Molarity of }Pb(NO_3)_2}$

$\text{Volume of }Pb(NO_3)_2=\frac{0.000525mole}{0.112mole/L}=4.6875\times 10^{-3}L=4.6875ml$

conversion used : (1 L = 1000 ml)

Therefore, the volume of $Pb(NO_3)_2$ needed are, 4.6875 ml

Answer 2

0.105 x 10 /2 / 0.112= 4.6875 ml of 0.112M Pb(NO3)2