How many grams of boron oxide, B2O3, can be produced from 20.0 g diborane (B2H6)?   B2H6 (l) + 3O2 (g) → B2O3 (s) + 3H2O (l)?

Molar mass of B₂H₆ = (10.8×2 + 1.0×6) g/mol = 27.6 g/mol
Molar mass of B₂O₃ = (10.8×2 + 16.0×3) g/mol = 69.6 g/mol

B₂H₆(ℓ) + 3O₂(g) → B₂O₃(s) + 3H₂O(ℓ)
Mole ratio B₂H₆ : B₂O₃ = 1 : 1

Moles of B₂H₆ reacted = (20.0 g) / (27.6 g/mol) = 0.7246 mol
Moles of B₂O₃ produced = 0.7246 mol
Mass of B₂O₃ produced = (0.7246 mol) × (69.6 g/mol) = 50.4 g

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OR:

(20.0 g B₂H₆) × (1 mol B₂H₆ / 27.6 g B₂H₆) × (1 mol B₂O₃ / 1 mol B₂H₆) × (69.6 g B₂O₃ / 1 mol B₂O₃)
= 50.4 g B₂O₃

20 g of boron oxide is how many moles?  (Divide 20 by the molecular weight of B2O3)

For every mole of B2O3, you get one mole of B2H6.  SO you now now how many moles of B2H6 you get,

Finally, multiply by the molecular weight of diborane to get your answer in grams.