Answer
0.215 g of AgCl will precipitate.
Work Step by Step
1. Calculate the amount of moles of each reactant:
$$30.0 \ mL \times \frac{1 \ L}{ 1000 \ mL} \times \frac{0.150 \ mol \ CaCl_2}{1 \ L} = 4.5 \times 10^{-3} mol$$
$$15.0 \ mL \times \frac{1 \ L}{1000 \ mL} \times \frac{0.100 \ mol \ AgNO_3}{1 \ L} = 1.5 \times 10^{-3} mol$$
2. Write the balanced equation:
$$CaCl_2(aq) + 2 \ AgNO_3(aq) \longrightarrow 2 \ AgCl(s) + Ca(NO_3)_2(aq)$$
$$4.5 \times 10^{-3} mol \ CaCl_2 \times \frac{2 \ mol \ AgCl}{1 \ mol \ CaCl_2} = 9.0 \times 10^{-3} mol \ AgCl$$
$$1.5 \times 10^{-3} mol \ AgNO_3 \times \frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3} = 1.5 \times 10^{-3} mol \ AgCl$$
$AgNO_3$ is the limiting reactant.
$ AgCl $ : ( 35.45 $\times$ 1 )+ ( 107.9 $\times$ 1 )= 143.4 g/mol
$$1.5 \times 10^{-3} mol \times \frac{143.4 \ g}{1 \ mol} = 0.215 \ g$$
