Correct Answer - A
Eq. (ii) multiplied by 3,
`2Fe(s)+(3)/(2)O_(2)(g)toFe_(2)O_(3)(s)" "(DeltaH=-193.4kJ)` . . .(i)
`3Mg(s)+(3)/(2)O_(2)(g)to3MgO(s)" "(DeltaH=-420.6kJ)` . . .(iii)
Subtracting eq. (i) from (iii),
`3Mg(s)+Fe_(2)O_(3)to3MgO+2Fe,DeltaH=-420.6-(-193.4)=-227.2kJ`