Question

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)

Answer 1

answer is 3.75L. I hope this helps :)

Answer 2

Answer: The volume of carbon dioxide produced will be 3.75 L

Step-by-step explanation:

STP conditions:

1 mole of as gas occupies 22.4 L of volume.

For the given chemical equation:

`6O_2(g)+C_6H_(12)O_6(g)\rightarrow 6H_2O(g)+6CO_2(g)`

As, glucose is present in excess, it is considered as excess reagent. Oxygen is considered as a limiting reagent.

By Stoichiometry of the reaction:

`6* 22.4L` of oxygen gas produces
`6* 22.4L` of carbon dioxide gas.

So, 3.75 L of oxygen gas will produce =
`(6* 22.4L)/(6* 22.4L)* 3.75L=3.75L` of carbon dioxide gas.

Thus, the volume of carbon dioxide produced will be 3.75 L