Question

25.00 mL of 0.500 M barium chloride solution is mixed with 25.00 mL of 0.500 silver nitrate solution. What mass of silver chloride will be formed? (How do you reach the conclusion that the answer is 1.79 g AgCl?)

Answer 1

moles of BaCl2 = 25 x 0.5 / 1000 = 0.0125

moles of AgNO3 = 25 x 0.5 / 1000 = 0.0125

BaCl2 + 2 AgNO3 ------------------------> Ba (NO3)2 + 2 AgCl (s)

1               2

0.0125     0.0125

here limiting reagent is AgNO3 . so the product based on limiting reagent.

2 mol AgNO3 -------------------------> 2 mol AgCl

0.0125 mol AgNO3 --------------------> 0.0125 mol AgCl

AgCl molar mass = 143.32 g/mol

mass = moles x molar mass

mass = 0.0125 x 143 .32

mass = 1.79 g

mass of silver chloride formed = 1.79 g