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Q.
When $22.4 \,L$ of $H _{2}( g )$ is mixed with $11.2\, L$ of $Cl _{2}( g )$, each at $STP$, the moles of $HCl (g)$ formed is equal to
AIPMTAIPMT 2014Some Basic Concepts of Chemistry
Solution:
The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
$\because 22.4 L$ volume at STP is occupied by
$Cl _{2}=1 $ mole
$\therefore 11.2\, L$ volume will be occupied by
$Cl _{2}=\frac{1 \times 11.2}{22.4} mol =0.5\, mol$
$22.4 \,L$ volume at STP is occupied by $H _{2}=1 \,mol$
Since, $Cl _{2}$ possesses minimum number of moles, thus it is the limiting reagent.
As per equation,
$1 $ mole of $ Cl _{2} =2 $ moles of $ HCl $
$\therefore 0.5 $ mole of $ Cl _{2} =2 \times 0.5 $ mole of $ HCl$
$=1.0 $ mole of $ HCl$
Hence, $1.0$ mole of $HCl (g)$ is produced by $0.5$ mole of $Cl _{2}$ [or $11.2 L$ ].
