One litre of oxygen at STP is made to react with three litres of carbon monoxide at STP. Which one is the limiting reactant?

Question

One litre of oxygen at STP is made to react with three litres of carbon monoxide at STP. Which one is the limiting reactant? (A) CO (B) O2 (C) CO2 (D) None of these

Tardigrade Admin · Accepted Answer

The chemical equation representing the reaction is

2 V o l 2 CO + 1 V o l O_{2} ⟶ 2 V o l 2 C O_{2}

Step 1. To calculate the volume and mass of

CO

(at STP) left unused after the reaction.
Applying Gay Lussac's law of gaseous volumes, 1 vol. of

O_{2}

reacts with

CO = 2

vol.

∴ 1 L

of

O_{2}

will react with

CO = 2 L

at STP
But the volume of

CO

taken

= 3 L

at STP

∴

Volume of CO (at STP) left unused

= 3 - 2 = 1 L

Now, by mole concept, mass

22.4 L

of

CO

at STP = Molecular mass (in

g

)

= 12 + 16 = 28 g

∴

Mass of

1 L

of

CO

at

STP = \frac{28}{22.4} \times 1 = 1.25 g

Thus, the mass of

CO

left unused

= 1.25 g

Step 2. To calculate the volume and mass of

C O_{2}

formed

1 L

of

O_{2}

at STP.
Applying Gay Lussac's law of gaseous volumes, 1 vol. of

O_{2}

produces

C O_{2} = 2 v o l

.

∴ 1 L

of

O_{2}

will produce

C O_{2} = 2 L

at STP
By mole concept, Mass of

22.4 L

of

C O_{2}

at STP

=

Molecular mass

(

in

g) = 12 + 2 \times 16 = 44 g

∴

Mass of

2 L

of

C O_{2}

at

STP = \frac{44}{22.4} \times 2 = 3.928 g

Thus, the mass of

C O_{2}

produced

= 3.928 g

Step 3. As oxygen has been completely used up, hence oxygen is the limiting reactant.

Q. One litre of oxygen at STP is made to react with three litres of carbon monoxide at STP. Which one is the limiting reactant?

CO

O2​

CO2​

None of these

$CO$

$O_{2}$

$C O_{2}$