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 :

When 22.4 litres of H_(2)(g) is mixied with 11.2 litres of Cl_(2)(g),

1.	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 :
Answer» 0.5 mol of HCl (g) 1.5 mol of HCl (g) 1.0 mol of HCl (g) 2.0 mol of HCl (g) SOLUTION :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.