Respuesta :
The question is incomplete. Complete question is attached below.
Solution:
The reaction involved in present electrochemical cell is:
Mg(s) + Zn2+(aq.) ↔ Zn(s) + Mg2+ (aq.)
Thus, from above reaction it can be seen that 1 mole of Zn2+ produces 1 mole of Mg2+.
Thus, 2.5 moles of Mg2+(aq) ions would be produced when 2.5 moles of Zn2+(aq) will react.
Solution:
The reaction involved in present electrochemical cell is:
Mg(s) + Zn2+(aq.) ↔ Zn(s) + Mg2+ (aq.)
Thus, from above reaction it can be seen that 1 mole of Zn2+ produces 1 mole of Mg2+.
Thus, 2.5 moles of Mg2+(aq) ions would be produced when 2.5 moles of Zn2+(aq) will react.

Answer: 2.5 moles of Mg+2
Since it is an example for intermolecular redox reaction. Here Mg gets oxidized to Mg+2 and Zn+2 gets reduced to Zn.
Mg(s)+ Zn+2(aq) -------> Mg+2(aq) + Zn(s)
From the balanced equation, 1:1 molar ratio of Mg and Zn+2. Hence 2.5 moles of Zn+2 produce 2.5 moles of Mg+2.