Calculating O2 Volume Required to Oxidize NO to NO2 at STP

What volume of O2 at STP is required to oxidize 8.0 L of NO at STP to NO2? What volume of NO2 is produced at STP?

Answer:

To oxidize 8.0 L of NO to NO2 at STP, 4.0 L of O2 is required, and the reaction will produce 8.0 L of NO2 at STP.

Explanation:

The question involves using stoichiometry to understand how gases react with each other, specifically how NO reacts with O2 to form NO2. At STP conditions (Standard Temperature and Pressure), equal volumes of gases will have the same number of molecules, according to Avogadro's hypothesis.

Since the balanced chemical equation for the reaction of NO with O2 is 2 NO(g) + O2(g) → 2 NO2(g), we can deduce that 1 mole of O2 is required for every 2 moles of NO, which implies that 4.0 L of O2 are required to oxidize 8.0 L of NO.

In return, 8.0 L of NO will produce 8.0 L of NO2, all at STP conditions.

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