Question

For the combustion of sucrose: C12H22O11 + 12O2 ---> 12CO2 + 11H2O there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?​

Answer 1

Answer:

For the combustion of sucrose: C12H22O11 + 12O2 ---> 12CO2 + 11H2O There are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent? Since the oxygen required is greater than that on hand, it will run out before the sucrose. Oxygen is the limiting reagent.

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