Question

240 mL of 0.32 M aluminum chloride reacts with excess silver nitrate in a double displacement reaction. What mass of silver chloride precipitate is produced?

AlCl3 + 3 AgNO3 -> Al(NO₃)₃ + 3 AgCl

Answer 1

The reaction is

$AlCl_3+3AgNO_3\Rightarrow Al{(NO_3)_2+3AgCl}$

The moles of Aluminium chloride =

$M=\dfrac{n_{AlCl_3}}{V(ml)}\times1000$

$0.32=\dfrac{n_{AlCl_3}}{240}\times1000$

$n_{AlCl_3}=0.077$ moles

Now according to the symmetry of the equation:

1 mole of Aluminium chloride gave 3 moles of silver chloride

Hence 0.077 moles of Aluminium chloride gave $3\times0.077=0.231$ moles of silver chloride.

Now mass of silver chloride =moles of silver chloride $\times$ molar mass of silver chloride

=$0.231\times143.32=33.10$ gm

So 33.10 gm of silver chloride is precipitated in produced after the reaction.