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?

-

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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they must be certified by the avma. ( a.)

mark me the branliest.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o