Answer : The volume of [tex]AgNO_3[/tex] required is, 60 L
Explanation :
First we have to calculate the moles of [tex]CaCl_2[/tex]
[tex]\text{Moles of }CaCl_2=\text{Concentration of }CaCl_2\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }CaCl_2=0.100M\times 3.00L=0.3mol[/tex]
Now we have to calculate the moles of [tex]AgNO_3[/tex]
The balanced chemical reaction will be:
[tex]2AgNO_3+CaCl_2\rightarrow 2AgCl+Ca(NO_3)_2[/tex]
From the balanced chemical reaction we conclude that,
As, 1 mole of [tex]CaCl_2[/tex] completely react with 2 moles of [tex]AgNO_3[/tex]
So, 0.3 mole of [tex]CaCl_2[/tex] completely react with [tex]0.3\times 2=0.6[/tex] moles of [tex]AgNO_3[/tex]
Now we have to calculate the volume of [tex]AgNO_3[/tex] required.
[tex]\text{Volume of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Concentration of }AgNO_3}[/tex]
[tex]\text{Volume of }AgNO_3=\frac{0.6mol}{0.0100M}=60L[/tex]
Therefore, the volume of [tex]AgNO_3[/tex] required is, 60 L
