A 295 g sample of crude nickel contains 80.0% Ni by mass, the rest being non-reacting impurity. It is treated with excess CO in the Mond process, and all the nickel is converted to $Ni (CO) X_{4}$ . Taking Ni = 59 and CO = 28, the mass of $Ni (CO) X_{4}$ formed and the volume of CO (at STP, 22.4 L/mol) consumed are respectively

Answer: C

Mass of Ni = 0.800 x 295 = 236 g, so n(Ni) = 236/59 = 4.0 mol.

Ni + 4 CO Ni (CO) X_{4}

gives 4.0 mol of carbonyl, M = 59 + 112 = 171 g/mol, mass = 684 g. CO used = 4 x 4.0 = 16.0 mol, volume = 358.4 L.