If 30.0 mL of 0.150 M CaC12 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgC1 precipitate?

2 answers

2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2

moles CaCl2 = M x L = ??
moles AgCl produced is 2x moles CaCl2 = ??
g AgCl = moles AgCl x molar mass AgCl.
1290.15grams