Answer
$ \Delta H^{\circ}_f = - 558.3 kJ/mol$
Work Step by Step
Equation for enthalpy of formation of BaO is
$Ba_{(s)} + O_{2(g)} → BaO_{(s)}$
According to stoichiometry, 1 mole of Ba is required to form BaO.
1 mole Ba = 137.327 g
2.740 g Ba at 298 K and 1 atm release 11.14 kJ heat.
Therefore heat released by 1 mole Ba (132.3 g) $= (11.14 \div 2.740) \times 137.327 = 558.3 kJ$
Hence enthalpy of formation of BaO is, $ \Delta H^{\circ}_f = - 558.3 kJ/mol$
