Question #d46e7

1 Answer
Oct 2, 2017

Answer:

#"MCl"_3#

Explanation:

The trick here is to realize that the number of moles of electrons needed to reduce #1# mole of #"M"^(?+)# cations will give you the net charge of the metal cations.

In your case, you know that #3# moles of electrons will reduce the #"M"^(?+)# cations to #1# mole of #"M"# metal, so you can say that

#"M"_ ((l))^(?+) + 3"e"^(-) -> "M"_ ((s))#

As you know, in any chemical reaction, charge must be conserved.

Since the metal is reduced to its elemental form, i.e. the charge on the metal will be #0#, you can say that you have

#(?+) + 3 * (-) = 0#

This implies that

#? = 3#

Therefore, your unknown chloride contained #3+# metal cations. Since chlorine forms #1-# anions, i.e. the chloride anion, #"Cl"^(-)#, you can say that the empirical formula of the chloride will be

#["M"]^(3+) + 3["Cl"]^(-) -> "MCl"_3#

This tells you that you need three #1-# chloride anions to balance the overall #3+# positive charge of a single #"M"^(3+)# cation.