The reaction is begun with 140.00 g of Si and 95.00 g of Cr2O3.
How many grams of the excess reactant are left after the reaction is complete?
(140.00 g Si) / (28.0855 g Si/mol) = 4.9847786 mol Si
(95.00 g Cr2O3) / (151.9904 g Cr2O3/mol) = 0.625039 mol Cr2O3
0.625039 mole of Cr2O3 would react completely with 0.625039 x (3/2) = 0.9375585 mole of Si, but there is more Si present than that, so Si is in excess.
((4.9847786 mol Si initially) – (0.9375585 mol Si reacted)) x (28.0855 g Si/mol) = 113.67 g Si left over