Đáp án:

$Mg;\ Ca\ Fe$

Giải thích các bước giải:

$Đặt\ n_{X}=4x(mol)\\ n_Y=2x(mol)\\ n_Z=x(mol)\\ M_X=3a(g/mol)\\ M_Y=5a(g/mol)\\ M_Z=7a(g/mol)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ X+H_2SO_4\to XSO_4+H_2\\ Y+H_2SO_4\to YSO_4+H_2\\ Z+H_2SO_4\to ZSO_4+H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{H_2}=n_X+n_Y+n_Z=7x(mol)\\ \Rightarrow 7x=0,35\\ \Rightarrow x=0,05\\ m_{\text{kim loại}}=11,6(g)\\ \Rightarrow 12ax+10ax+7ax=11,6\\ \Rightarrow 29ax=11,6\\ \Rightarrow 1,45a=11,6\\ \Rightarrow a=8\\ \Rightarrow \begin{cases}M_X=24(g/mol)\\M_Y=40(g/mol)\\M_Z=56(g/mol)\end{cases}\\ \Rightarrow \begin{cases}X:\ Mg\\Y:\ Ca\\Z:\ Fe\end{cases}$