What should be the weight and moles of AgCl precipitate obtained on adding 500 ml of 0.20 M HCl in 30 g of AgNO3 solution ?  AgNO3=170

AgNO3 +HCl  →  AgCl+HNO3               1mole  + 1mole→ 1molegiven,   30170moles+500×0.21000 →?                   =0.176   + 0.1molesHCl is the limiting reagentNumber of moles of AgCl formed =0.1×Molecular wt.=0.1×143.5=14.35g