Answer: 6.
Explanation:
1) Aluminum
[tex]Al^0-3e^----\ \textgreater \ Al^{3+}[/tex]
So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.
2) Manganesium
[tex]Mn^{2+}+2e^{-}---\ \textgreater \ Mn[/tex]
So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.
3) Balance
Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:
[tex]2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}
3Mn^{2+}+6e^{-}---\ \textgreater \ 3Mn^{0}[/tex]
4) Net equation
Add the two half-equations:
[tex]2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}[/tex]
As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.
The right side has 2 Al, 3 Mn, and 2*3 positive charges.
So, the equation is balanced.
5) Count the number of electrons involved.
As you see 2 atoms of aluminum lost 6 electrons (3 each).
That is the answer to the question. 6 electrons will be lost.
