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 molreaction 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=1Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is formed.