Question #d46e7

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Question #d46e7

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Answer 1 · 2017-10-02T13:45:06

${MCl}_{3}$ $"MCl"_3$

Explanation:

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

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

$M_{(l)}^{? +} + 3 e^{-} \to M_{(s)}$ $"M"_ ((l))^(?+) + 3"e"^(-) -> "M"_ ((s))$

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 $0$ $0$ , you can say that you have

$(? +) + 3 \cdot (-) = 0$ $(?+) + 3 * (-) = 0$

This implies that

$? = 3$ $? = 3$

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

${[M]}^{3 +} + 3 {[Cl]}^{-} \to {MCl}_{3}$ $["M"]^(3+) + 3["Cl"]^(-) -> "MCl"_3$

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

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Question #d46e7

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