How much nitrogen is needed to interact with 3.75 liters of oxygen?

Let’s write the reaction of nitrogen with oxygen:
N2 + O2 = 2NO
n (O2) = 3.75 / 22.4 = 0.1674 mol
Nitrogen reacts as much as oxygen, that is, n (N2) = 0.1674 mol
We calculate the volume: V (N2) = 0.1674 * 22.4 = 3.75 l.
To interact with 3.75 liters of acid, the same volume of nitrogen is required.

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