When 22.4 litres of `H_(2)(g)` is mixied with 11.2 litres of `Cl_(2)(g)`, each at S.T.P, the moles of HCl (g) formed is equal to :

Options

A. 0.5 mol of HCl (g)
B. 1.5 mol of HCl (g)
C. 1.0 mol of HCl (g)
D. 2.0 mol of HCl (g)

Answer

C

Explanation

No. of moles in 22.4 L of `H_(2)(g)` at S.T.P = 1 mol
No. of moles in 11.2 L of `Cl_(2)(g)` at S.T.P = 0.5 mol
`underset("1 mol")(H_(2)(g))+underset("1 mol")(Cl_(2)(g))rarrunderset("2 mol")(2HCl(g))`
`Cl_(2)(g)` is the limiting reactant since it has only 0.5 mole available for chemical reaction
`:.` No. of moles of HCl (g) formed = 1mol.