1 lit. of oxygen at STP is made to react with 3 lit. of Carbon Monoxide at STP. Calculate the mass of each substance found after the reaction. Which one is he limiting reactant?

2CO + O₂ → 2CO₂

1 mole O₂ needs 2 mole CO to form 2 moles CO₂

At STP one mole of an ideal gas occupies 22.4 l.

So 1 l O₂ occupies 1/22.4 = 0.0446 moles

and 3 l CO occupies  3/22.4 = 0.1339 moles

To find out the limiting reagent divide the no of moles of reactants with their stoichiometric coefficient.
For O₂: 0.0446
For CO: 0.1339/2 = 0.0669
The lowest number indicates the limiting reagent which in this case is oxygen.

Molecular mass of CO =28 g/mol and CO₂ = 44 g/mol. 
In the given reaction,

1) 0.0446 moles of O₂ will consume 2*0.0446 mole CO.
CO remaining unreacted = 0.1339 -2*0.0446
0.0447 moles CO or 0.0447*28 = 1.2516 gm CO


2) 0.0446 moles O₂ will produce 2*0.0446 CO₂​.
Amount of CO₂ = 2*0.0446*44 = 3.9248 gm CO₂

Hence, there will be 1.2516 gm unreacted CO and 3.9248 gm CO₂ after the reaction between given O₂ and CO.