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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)

Respuesta :

answer is 3.75L. I hope this helps :)

Answer: 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:

[tex]6O_2(g)+C_6H_{12}O_6(g)\rightarrow 6H_2O(g)+6CO_2(g)[/tex]

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

By Stoichiometry of the reaction:

[tex]6\times 22.4L[/tex] of oxygen gas produces [tex]6\times 22.4L[/tex] of carbon dioxide gas.

So, 3.75 L of oxygen gas will produce = [tex]\frac{6\times 22.4L}{6\times 22.4L}\times 3.75L=3.75L[/tex] of carbon dioxide gas.

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

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