Explanation:
a) Since this is a double displacement reaction, we write the balanced equation as
[tex]2AgNO_3(aq) + CaCl_2(aq) \\ \rightarrow 2AgCl(s) + Ca(NO_3)_2(aq)[/tex]
b) Next we find the number of moles of AgNO3 in the solution.
[tex](0.005\:\text{L})(0.500\:M\:AgNO_3) \\ = 0.0025\:\text{mol}\:AgNO_3[/tex]
Next, use the molar ratio to find the necessary amount of CaCl2 to react with the AgNO3:
[tex]0.0025\:\text{mol}\:AgNO_3× \left(\dfrac{1\:\text{mol}\:CaCl_2}{2\:\text{mol}\:AgNO_3} \right)[/tex]
[tex]= 0.00125\:\text{mol}\:CaCl_2[/tex]
The volume of 0.500 M solution of CaCl2 necessary to react all of the given AgNO_3 is then
[tex]V = \dfrac{0.00125\:\text{mol}\:CaCl_2}{0.500\:\text{M}\:CaCl_2}[/tex]
[tex]= 0.0025\:\text{L} = 2.5\:\text{mL}\:CaCl_2[/tex]
c) The theoretical yield can then be calculated as
[tex]0.0025\:\text{mol}\:AgNO_3 × \left(\dfrac{2\:\text{mol}\:AgCl}{2\:\text{mol}\:AgNO_3} \right)[/tex]
[tex]= 0.0025\:\text{mol}\:AgCl[/tex]
Converting this amount of AgCl into grams, we get
[tex]0.0025\:\text{mol}\:AgCl × \left(\dfrac{143.32\:\text{g}\:AgCl}{1\:\text{mol}\:AgCl} \right)[/tex]
[tex]= 0.358\:\text{g}\:AgCl[/tex]
