The equation represents the combustion of sucrose (C12H22O11). C12H22O11 + 12O2 12CO2 + 11H2OIf there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction?

Question

Revolt · Accepted Answer

0.029 mol

Limiting Reactant and Percent Yield

The equation represents the combustion of sucrose (C12H22O11). C12H22O11 + 12O2 12CO2 + 11H2OIf there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction?

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