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4NH3(g) + 5O2(g) -> 4NO(g) +6H20(g) If 3.5 L of oxygen gas at STP react with an excess amount of ammonia, how many liters of nitrogen monoxide will be produced

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Answer:

[tex]V_{NO}=2.8L[/tex]

Explanation:

Hello!

In this case, we consider the given reaction to realize there is a 5:4 mole ratio between oxygen and nitrogen monoxide; thus, we infer that, at STP conditions, such mole ratio is eligible as a volume ratio too; therefore, the produced liters of nitrogen monoxide gas is:

[tex]V_{NO}=3.5LO_2*\frac{4LNO}{5LO_2} \\\\V_{NO}=2.8L[/tex]

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