When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P, the moles of HCl (g)formed is equal to
1 mole ≡22.4 litres at S.T.P.
nH2=22.422.4=1 mol,nCl2=11.222.4=0.5 mol
reaction is as
H2(g) + Cl2(g) ⟶ 2HCl(g)
Initial 1 mol0.5 mol 0 Final (1-0.5)=0.5 (0.5-0.5)=0 2×0.5=1
Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is formed.