Question

1litre of O_{​​​​​​2} at STP is made to react with 3litre of carbon monoxide at STP . Calculate the mass of each substance formed after the reaction?

Answer 1

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.