When `22.4 L` of `H_(2)(g)` is mixed with 11.2 of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal to

Options

A. 1 mole of HCl (g)
B. 2 moles of HCl (g)
C. 0.5 mole of HCl (g)
D. 1.5 moles of HCl (g)

Answer

A

Explanation

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.
`{:(,H_(2)(g),+,Cl_(2)(g),to,2HCl(g)),("Initial vol.",22.4L,,11.2L,,2 "mol"):}`
`:'`22.4 L volume at STP is occupied by
`Cl_(2)=1` mole
`:.` 11.2L volume will be occupied by
`Cl_(2)=(1xx11.2)/(22.4)"mol"=0.5"mol"`
22.4 L volume at STP is occupied by `H_(2)=1` mol
Thus , `{:(H_(2)(g),+,Cl_(2)(g),to,2HCl(g)),(1"mol",,0.5"mol",,):}`
Since , `Cl_(2)` possesses minimum number of moles thus it is limiting reagent.
As per equation
1 mole of `Cl_(2)=2` moles of HCl
`:.` 0.5 mole of `Cl_(2)=2xx0.5` 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 11.2L].