What is the maximum mass of chromium, Cr, that can be extracted from 76g of chromium (III) oxide?

Method 1:

Mass of 1 mole of Cr₂O₃ = (52×2 + 16×3) g = 152 g
Mass of Cr in 1 mole of Cr₂O₃ = 52×2 g/mol = 104 g
Mass fraction of Cr in Cr₂O₃ = 104/152

Maximum mass of Cr extracted from 76 g of Cr₂O₃
= 76 × (104/152) g
= 52 g

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Method 2:

Balanced equation for the reaction:
2Cr₂O₃ → 4Cr + 3O₂
Mole ratio Cr₂O₃ : Cr = 2 : 4

(76 g Cr₂O₃) × (1 mol Cr₂O₃ / 152 g Cr₂O₃) × (4 mol Cr / 2 mol Cr₂O₃) × (52 g Cr / 1 mol Cr)
= 52 g Cr

use dimensional analysis and mass fraction as your unit factor.
.. 76g Cr2O3 * (104g Cr / 152g Cr2O3) = 52g

Cr is 68.42% of Cr2O3 

so the maximium recovery of the metal = 68.42% of 76g = 52g