# Answer to Question #191707 in General Chemistry for Ayame

Question #191707

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

1
2021-05-12T06:30:29-0400

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.

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