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Alexxx [7]
3 years ago
15

At STP, when 3.75 liters of oxygen reacts with excess glucose, what volume of carbon dioxide does it produce?

Physics
2 answers:
I am Lyosha [343]3 years ago
5 0
Answer is 3.75L. I hope this helps :)

iragen [17]3 years ago
5 0

<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:

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\times 22.4L of oxygen gas produces 6\times 22.4L of carbon dioxide gas.

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

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

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3 years ago
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valentinak56 [21]

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