Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. what volume of nitrogen monoxide would be produced by this reaction if of oxygen were consumed?

The question is incomplete.

The probable question is

Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. what volume of nitrogen monoxide would be produced by this reaction if 6.3 L of oxygen were consumed?

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5.04 L

Explanation: -

The balanced chemical equation for this process is

4NH₃ + 5O₂ → 4NO + 6H₂O

Using gay lussac law of combining volumes from the balanced chemical equation we see

5 volumes of O₂ react to give 4 volumes of NO

6.3 L of oxygen will give x 6.3 L

= 5.04 L of NO

The volume of NO produced by 6.3 L of oxygen is 5.04 L.

Explanation:

The question states the formation of water vapors and nitrogen monoxide in the reaction of Ammonia with Oxygen.

If the amount of oxygen consumed  = 6.3 L, the volume of NO will be?

This implies that,

5 moles of oxygen produces 4 moles of NO.

So, 6.3 L of Oxygen produces = 6.3 L

NO produced by 6.3 L of oxygen = 5.04 L.

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