120 g of CaO reacts with 32 g of ${\mathrm{SO}}_2$ To produce CaSO₃. The mass of ${\mathrm{CaSO}}_3$ Produced is ( Atomic mass of Ca and S are 40 and 32 respectively)

$\underset{2.7}{\mathrm{CaO}} + \underset{0.5}{{\mathrm{SO}}_2} \to {\mathrm{CaSO}}_3$

${\mathrm{SO}}_2$ is limiting reagent 

moles of ${\mathrm{CaSO}}_3$=0.5

Now,

1 mole of ${\mathrm{SO}}_2$ is required to make 1 mole of ${\mathrm{CaSO}}_3$.

So,0.5 of ${\mathrm{SO}}_2$ is required to make 0.5 moles of ${\mathrm{CaSO}}_3$.

Therefore, mass of ${\mathrm{CaSO}}_3$ produce= no. of moles of ${\mathrm{CaSO}}_3$ $\times$ its molecular mass

$=0.5\times 120 = 60\mathrm{g}$

Therefore, the correct option is (D).