If there are 10.0 g of sucrose 8.0 g of oxygen how many moles of sucrose are available for this reaction

Answer 1

Write a correctly balanced equation:

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

moles sucrose present = 10.0 g x 1 mole/342 g = 0.0292 moles

moles O₂ present = 8.0 g x 1 mole/32 g = 0.25 moles

Limiting reactant is O₂ since it takes 12 moles O₂ per mole sucrose

Thus moles of sucrose available for reaction = 0.25 moles O₂ x 1 mole sucrose/12 moleO₂ = 0.021 moles