Câu 7:
$n_{H_2}=\dfrac{4,958}{22,4}=0,221339(mol)$
$X+2HCl\to XCl_2+H_2$
$\to n_X=n_{H_2}=0,221339(mol)$
$\to M_X=\dfrac{13}{0,221339}=58,73$ (bạn xem lại đề)
Câu 8:
$Hn_{H_2SO_4}=0,3(mol)$
$Z_2O_3+3H_2SO_4\to Z_2(SO_4)_3+3H_2O$
$\to n_{Z_2O_3}=\dfrac{n_{H_2SO_4}}{3}=0,1(mol)$
$\to M_{Z_2O_3}=\dfrac{16}{0,1}=160$
$\to 2M_Z+16.3=160$
$\to M_Z=(160-16.3):2=56(Fe)$
Vậy $Z$ là sắt