Mg(s) + ½O2(g) → MgO(s) + 601 kJ
\(\Delta H = -601\mathrm{kJ/mol}\)
Vi tecknar den frigjorda energin \(q\):
\(q = \Delta H \cdot n_\mathrm{Mg}\)
\[n_\mathrm{Mg} = \frac {m_\mathrm{Mg}}{M_\mathrm{Mg}} = \frac {1,22\mathrm{g}}{24,31\mathrm{g/mol}} = 0,05018511\mathrm{mol}\]
\(q = -601\mathrm{kJ/mol} \cdot 0,05018511\mathrm{mol} = -30,1612505\mathrm{kJ} ≈ -30,2\mathrm{kJ}\)
Det frigörs sålunda 30,2kJ.
