Answer:
Choice A. [tex]x[/tex] electrons would be required for displacing [tex]9\; \rm g[/tex] of aluminum from a solution of [tex]\rm Al^{3+}[/tex] ions.
Assumption: by "[tex]\rm Ag[/tex] ions" the question meant [tex]\rm Ag^{+}[/tex] with a charge of [tex]+1[/tex] on each ion.
Explanation:
The question states that the relative atomic mass of [tex]\rm Ag[/tex] is [tex]108[/tex]. In other words, each mole of
Therefore, that [tex]108\; \rm g\![/tex] of silver that were formed would contain [tex]1\; \rm mol[/tex] of silver atoms.
Metallic silver would precipitate out of this [tex]\rm Ag^{+}[/tex] solution only after these ions are turned into [tex]\rm Ag[/tex] atoms.
One [tex]\rm Ag^{+}[/tex] ion carries one unit of positive electrical charge. On the other hand, each [tex]e^{-}[/tex] carries one unit of negative electrical charge.
Therefore, each [tex]\rm Ag^{+}\![/tex] ion will need to gain one electron to form a neutral [tex]\rm Ag[/tex] atom.
[tex]{\rm Ag^{+}}\; (aq) + e^{-} \to {\rm Ag}\; (s)[/tex].
At least [tex]1\; \rm mol[/tex] of electrons would be required to turn [tex]1\; \rm mol\![/tex] of [tex]\rm Ag^{+}[/tex] ions into that [tex]1\; \rm mol\!\![/tex] of silver atoms (which have a mass of [tex]108\; \rm g\![/tex].)
Hence, [tex]x = 1\; \rm mol[/tex].
Unlike [tex]\rm Ag^{+}[/tex] ions, each aluminum ion [tex]\rm Al^{3+}[/tex] carries three units of positive electrical charge. That is three times the amount of charge on one [tex]\rm Ag^{+}\![/tex] ion. Therefore, three electrons will be required to turn one [tex]\rm Al^{3+}\![/tex] ion to an [tex]\rm Al[/tex] atom.
[tex]{\rm Al^{3+}}\; (aq) + 3\, e^{-} \to {\rm Al}\; (s)[/tex]
The question states that the relative atomic mass of [tex]\rm Al[/tex] is [tex]27[/tex]. Therefore, each mole of [tex]\rm Al\![/tex] atoms would have a mass [tex]27\; \rm g[/tex]. There would be [tex]\displaystyle \frac{9\; \rm g}{27\; \rm g \cdot mol^{-1}} = \frac{1}{3} \; \rm mol[/tex] of atoms in that [tex]9\; \rm g[/tex] of [tex]\rm Al\!\![/tex].
It takes [tex]3\; \rm mol[/tex] of electrons to turn one mole of [tex]\rm Al^{3+}[/tex] ions to one mole of [tex]\rm Al[/tex] atoms. Hence, [tex]\displaystyle \frac{1}{3}\times 3\; \rm mol = 1\; \rm mol[/tex] of electrons would be required to produce that [tex]\displaystyle \frac{1}{3}\; \rm mol[/tex] of [tex]\rm Al\![/tex] atoms (which has a mass of [tex]9\; \rm g[/tex]) from [tex]\rm Al^{3+}\![/tex] ions.
That corresponds to the first choice, [tex]x[/tex] electrons.
