Answer:
[tex]V_{NO}=2.8L[/tex]
Explanation:
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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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