Answer
0.21525 g AgCl
Work Step by Step
First, we calculate the number of moles of $CaCl_{2}$ and $AgNO_3$.
$C_M$ = $\frac{n}{V}$
$n_{CaCl_2}$ = 0.15$\times$0.03 = 0.0045 moles
$n_{AgNO_3}$ = 0.1$\times$0.015 = 0.0015 moles
We notice there is an excess of $CaCl_2$, so the entire quantity of $AgNO_3$ is consumed
$M_{AgCl}$ = 108+35.5 = 143.5 g/mole
The reaction that takes place is:
$CaCl_{2}$ + 2$AgNO_3$ = $Ca(NO_3)_2$ + 2AgCl
0.0015 moles $AgNO_3$.................. x g AgCl
2 moles $AgNO_3$..................2*143.5 g AgCl
=> x = $\frac{2*143.5*0.0015}{2}$ = 0.21525 g AgCl
