\(m_X = m_{kim\ loại} + m_{O}\\ \Rightarrow n_O = \dfrac{9,2-7,6}{16} = 0,1(mol)\\ n_{H_2} = \dfrac{0,672}{22,4} = 0,03(mol)\\ 2H^+ + 2e \to H_2\\ 2H^+ + O^{2-} \to H_2O\\ n_{H^+} = 2n_{H_2} + 2n_O = 0,1.2 + 0,03.2=0,26(mol)\\ n_{H_2SO_4} = \dfrac{1}{2}n_{H^+} = 0,13(mol)\\ \Rightarrow V = \dfrac{0,13}{0,5} = 0,26(lít)\)