Answer
See answer below.
Work Step by Step
a) From the standard states:
$2\ Cr(s)+3/2\ O_2(g)\rightarrow Cr_2O_3(s)$
Enthalpy of formation:
$-1134.7\ kJ/mol$
b) Number of moles of Cr:
$2.4\ g\div 51.996\ g/mol=0.046\ mol$
Enthalpy:
$0.046\ mol\times 1/2\times-1134.7\ kJ/mol= -26.2\ kJ$
