The chemical equation representing the reaction of silver nitrate with barium chloride:
[tex] 2AgNO_{3}(aq) + BaCl_{2}(aq)--> 2AgCl (s) + Ba(NO_{3})_{2}(aq) [/tex]
Given mass of barium chloride = 4.62 g
Moles of [tex] BaCl_{2} = 4.62 g BaCl_{2}*\frac{1 mol BaCl_{2}}{208.23 g BaCl_{2}} = 0.0222 mol BaCl_{2} [/tex]
Moles of AgCl = [tex] 0.0222 mol BaCl_{2} * \frac{2 mol AgCl}{1 mol BaCl_{2}} [/tex] = 0.0444 mol AgCl
Mass of AgCl = [tex] 0.0444 mol AgCl * \frac{143.32 g AgCl}{1 mol AgCl} = 6.36 g AgCl [/tex]
