The 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?

0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles