The correct option is D 9 g of the metal can completely react with 8 g of oxygen
Since 5.6 litre will be occupied by 14 mole of NTP, so the molecular weight of metal chloride
=33.375×4=133.5 g
Let, the formula of the chloride be MClx
x × mole of metal = 1× mole of Cl
Let the total mass of metal chloride = 100 g
Then mass of Cl = 80 g
& mass of metal = 20 g
& let molecular weight of metal=MA
x×20MA=1×8035.5Also,MA+35.5x=133.5
On solving, x = 3 & MA=27 g
Equivalent mass of the metal = 9 g
Hence, 9 g of the metal can combine with 8 g oxygen.