Question

What amount of silver chloride will form when 100.0 mL of 0.400 M silver nitrate is reacted with 90.0 mL of 0.250 M calcium chloride?

Answer 1

moles of AgNO3 =100 x 0.4 / 1000 = 0.04

moles of CaCl2 = 90 x 0.250 / 1000 = 0.0225

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

2 mole          1mole

0.04 mol     0.0225

limiting reagent is AgNO3 .

so moles of AgCl formed = 0.04

mass of AgCl = moles x molar mass

                     = 0.04 x 143.3

                    = 5.73 g

amount of AgCl = 5.73 g