Answer: 11.8 grams of [tex]CaCl_2[/tex] is required.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]AgNO_3[/tex] solution = 0.506 M
Volume of solution = 420 mL
Putting values in equation 1, we get:
[tex]0.506M=\frac{\text{Moles of} AgNO_3\times 1000}{420ml}\\\\\text{Moles of }AgNO_3=\frac{0.506mol/L\times 420}{1000}=0.212mol[/tex]
The balanced chemical reaction is:
[tex]2AgNO_3+CaCl_2\rightarrow 2AgCl+Ca(NO_3)_2[/tex]
2 mole of [tex]AgNO_3[/tex] requires = 1 mole of [tex]CaCl_2[/tex]
0.212 moles of [tex]AgNO_3[/tex] require = [tex]\frac{1}{2}\times 0.212=0.106[/tex] moles of [tex]CaCl_2[/tex]
Mass of [tex]CaCl_2[/tex] =[tex]moles\times {\text {Molar Mass}}=0.106mol\times 111g/mol=11.8g[/tex]
