For the given reaction, what volume of NO2Cl can be produced from 3.3 L of Cl2 , assuming an excess of NO2 ? Assume the temperature and pressure remain constant. 2NO2(g)+Cl2(g)⟶2NO2Cl(g)

Answer 1

2NO₂(g) + Cl₂(g) ==> 2NO₂Cl(g) ... balanced equation

This tells us that for every 1 mole of Cl₂, you can produce 2 mols of NO₂Cl

It also tells us that for every 1 liter of Cl₂, you can produce 2 liters of NO₂Cl

3.3L Cl₂ x 2 L NO₂Cl / 1 L Cl₂ = 6.6 L NO₂Cl can be formed