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)?

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huy219 huy219 · Answer 1 · 2018-05-09T14:51:14+02:00

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

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Alleei Alleei · Answer 2 · 2019-08-13T09:16:31+02:00

Answer : The volume of [tex]Pb(NO_3)_2[/tex] needed are, 4.6875 ml

Explanation :

First we have to calculate the moles of [tex]KI[/tex].

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

Now we have to calculate the moles of [tex]Pb(NO_3)_2[/tex]

The given balanced chemical reaction is,

[tex]Pb(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+PbI_2[/tex]

From the balanced chemical reaction, we conclude that

As, 2 moles of KI react with 1 mole of [tex]Pb(NO_3)_2[/tex]

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

Now we have to calculate the volume of [tex]Pb(NO_3)_2[/tex]

[tex]\text{Volume of }Pb(NO_3)_2=\frac{\text{Moles of }Pb(NO_3)_2}{\text{Molarity of }Pb(NO_3)_2}[/tex]

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

conversion used : (1 L = 1000 ml)

Therefore, the volume of [tex]Pb(NO_3)_2[/tex] needed are, 4.6875 ml