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 limiting reacgent gives the moles og product formed
in the reaction.
`underset(22.4L)(H_(2)(g)) + underset(11.2 L)(Cl_(2) (g)) rarr underset(2" mol")(2HCl(g))`
`because 22.4` volume at STP is occupied by
`therefore 11. 2` L volume will be occupied by
`Cl_(2) = (1xx11.2)/22.4 "mol"=0.5` mol
Thus, `underset(1" mol")(H_(2)(g))+underset(0.5" mol")(Cl_(2)(g)) rarr 2HCl(g)`
Since, `Cl_(2)` possesses minimum number of moles, thus
it is the limiting reagent.
As per equation , mol `Cl_(2)=2" mol HCl"`
`therefore 0.5 ` mol `Cl_(2) = 2 xx 0.5` mol HCl
= `1.0` mol HCl
Hence,`1.0` mole of HCl (g) is produced by `0.5` mole of
`Cl_(2) ` (or `11.2`L).