<u>Answer:</u> The volume of carbon dioxide produced will be 3.75 L
<u>Explanation:</u>
STP conditions:
1 mole of as gas occupies 22.4 L of volume.
For the given chemical equation:
As, glucose is present in excess, it is considered as excess reagent. Oxygen is considered as a limiting reagent.
By Stoichiometry of the reaction:
of oxygen gas produces of carbon dioxide gas.
So, 3.75 L of oxygen gas will produce = of carbon dioxide gas.
Thus, the volume of carbon dioxide produced will be 3.75 L