Question

Question #d46e7

Answer 1

`"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.