iron(III)oxide +zinc metal..react to produce iron metal + zinc(II)oxide[25g each reactant]...product produced=?

Fe2O3 + 3 Zn ? 2 Fe + 3 ZnO

(25 g Fe2O3) / (159.6887 g Fe2O3/mol) = 0.15655 mol Fe2O3
(25 g Zn) / (65.4094 g Zn/mol) = 0.38221 mol Zn

0.15655 mole of Fe2O3 would react completely with 0.15655 x 3 = 0.46965 mole of Zn, but there is not that much Zn present, so Zn is the limiting reactant.

(0.38221 mol Zn) x (2/3) x (55.8452 g Fe/mol) = 14 g Fe produced
(0.38221 mol Zn) x (3/3) x (81.4088 g ZnO/mol) = 31 g ZnO produced
(0.38221 mol Zn) x (1/3) = 0.1274 mol Fe2O3 reacted
(0.15655 mol Fe2O3 initially - 0.1274 mol Fe2O3 reacted) x (159.6887 g Fe2O3/mol) = 4.7 g Fe2O3 left over.
There is no Zn left over.