C12H22O11 + 12O2 - --> 12CO2 + 11H2O

there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?

Oxygen is the limiting reactant.

Explanation:

Based on the reaction:

C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.

10.0g of sucrose (Molar mass: 342.3g / mol) are:

10.0g C₁₂H₂₂O₁₁ * (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁

And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:

10.0g O₂ * (1mole / 32g) = 0.3125 moles of O₂

For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):

0.0292 moles of C₁₂H₂₂O₁₁ * (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = 0.3504 moles of O₂

As you have just 0.3125 moles of O₂, oxygen is the limiting reactant.