He equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction?

The moles of sucrose that are available for this reaction is 0.0292 moles

Explanation:

Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O

This combustion is: C₁₂H₂₂O₁₁ + 12O₂  → 12CO₂ + 11H₂O

We have to conver the mass to moles, to find out the limiting reactant

10 g . 1 mol / 342 g = 0.0292 moles of sucrose

8 g . 1mol / 32g = 0.250 moles of O₂

The moles of sucrose that are available for this reaction is 0.0292 moles

Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.

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