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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?
Some Basic Concepts of Chemistry
Solution:
The chemical equation representing the reaction is $\underset{2 Vol }{2 CO }+\underset{1 Vol }{ O _2} \longrightarrow \underset{2 Vol }{2} CO _2$ 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.
$\therefore 1\, L$ of $O _2$ will react with $CO =2 \,L$ at STP
But the volume of $CO$ taken $=3\, L$ at STP
$\therefore$ 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 $g$ ) $=12+16=28\, g$
$\therefore$ 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 $CO _2$ formed $1\, L$ of $O _2$ at STP.
Applying Gay Lussac's law of gaseous volumes, 1 vol. of $O _2$ produces $CO _2=2 vol$.
$\therefore 1\, L$ of $O _2$ will produce $CO _2=2\, L$ at STP
By mole concept, Mass of $22.4 \,L$ of $CO _2$ at STP
$=$ Molecular mass $($ in $g )=12+2 \times 16=44 \,g$
$\therefore$ Mass of $2\, L$ of $CO _2$ at $STP =\frac{44}{22.4} \times 2=3.928\, g$
Thus, the mass of $CO _2$ produced $=3.928\, g$ Step 3. As oxygen has been completely used up, hence oxygen is the limiting reactant.