For the given reaction, what volume of O2 would be required to react with 4.8 L of NO , measured at the same temperature and pressure? 2NO(g)+O2(g)⟶2NO2(g)

Answer 1

2NO(g) + O₂(g) ==> 2NO₂(g) ... balanced equation

From this balanced equation we see that 2 moles NO reacts with 1 mole O₂. At constant pressure and temperature, we can also say that 2 liters of NO reacts with 1 liter of O₂. Then we can determine volume of O₂ needed to react with 4.8 L of NO as follows:

4.8 L NO x 1 L O₂ / 2 L NO = 2.4 L of O₂ required