\[{n_{{H_2}}} = \frac{{1,456}}{{22,4}} = 0,065(mol)\]

\[\begin{array}{l}2\mathop H\limits^{ + 1} + 2e \to {\mathop H\limits^0 _2}\\{\mathop O\limits^0 _2} + 4e \to 2\mathop O\limits^{ - 2} \\{n_{{O_2}}} = \frac{{2.0,065}}{4} = 0,0325(mol)\end{array}\]

m_oxit = \[\frac{{2,29}}{2} + 0,0325.32 = 2,185(g)\]