Final answer:
Glucose is the limiting reagent when 25 grams of glucose reacts with 40 grams of oxygen in cellular respiration. There are 0.416 moles, or 13.32 grams, of oxygen in excess after the reaction.
Step-by-step explanation:
The student's question requires identifying the limiting reagent in cellular respiration when 25 grams of glucose reacts with 40 grams of oxygen. The chemical equation for cellular respiration is C6H12O6 + 6O2 → 6CO2 + 6H2O + Energy. To find the limiting reagent, we need to use molar masses and mole ratios.
Firstly, calculate the moles of both reactants. The molar mass of glucose (C6H12O6) is 180.16 g/mol and for oxygen (O2) it is 31.9988 g/mol.
Number of moles of glucose = 25 g / 180.16 g/mol ≈ 0.139 moles.
Number of moles of oxygen = 40 g / 31.9988 g/mol ≈ 1.250 moles.
From the equation, we can see that 1 mole of glucose reacts with 6 moles of oxygen. Therefore, for 0.139 moles of glucose, we would need 0.139 * 6 = 0.834 moles of oxygen. Since we have more oxygen than required (1.250 moles), glucose is the limiting reagent and oxygen is in excess. To determine the excess amount of oxygen, subtract the required amount from the available amount: 1.250 moles - 0.834 moles = 0.416 moles in excess, which is equivalent to 0.416 moles * 31.9988 g/mol = 13.32 grams of oxygen.