At stp, when 3.75 liters of oxygen reacts with excess glucose, what volume of carbon dioxide does it produce? 6o2(g) + c6h12o6(g) → 6h2o(g) + 6co2(g)

The volume of carbon dioxide produced will be 3.75 L

Explanation:

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

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