When 25mL of a 0.5 M solution of ZnCl2 reacts with an excess of magnesium, how much zinc metal is produced?

Answer 1

Mg(s) + ZnCl2(aq) ==> Zn(s) + MgCl2(aq)

moles ZnCl2 present = 25 ml x 1 L/1000 ml x 0.5 mol/L = 0.0125 moles ZnCl2

moles Zn(s) produced = 0.0125 mol ZnCl2 x 1 mol Zn / mol ZnCl2 x 65.4 g Zn/mol Zn = 0.8 g Zn produced