Two moles each of hydrogen and chlorine(at STP) produce ‘V’ litres of HCl. ‘V’ litres of this HCl can exactly neutralize ‘W’ grams of Na2CO3. Value of ‘W’ is

1 H2g+Cl2g→2HClg ;   1 mole    1 mole        44.8l 2 moles    2 moles      89.6l 2 Na2CO3+2HCl→2NaCl+H2O+CO2 ;       106 g       2×22.4 l 2×106 g             89.6 l