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. 0.5mole of HCl (g)
D. 1.5 mole of HCl(g)

Answer

B

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, identity 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 reation.
`underset(22.4L)(H_(2)(g)) + underset(11.2L)(cl_(2)(g)) to underset("2 mol")(2HCl(g))`
`because` 22.4 volume at STP is occupied by
`Cl_(2) = 1 "mole"`
`therefore ` 11.2L volume will be occupoied by
`Cl_(2) = (1 xx 11.2)/(22.4)" mole" = 0.5 mol`
Thus `underset(1 mol)(H_(2)(g)) + underset(0.5 mol)(Cl_(2)(g)) to 2HCl (g)`
Since `Cl_(2)` possesses minimum number of moles , thus it is the limiting reagent
As per equation , `mol Cl_(2) = 2mol` 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 11.2L)`