Question

Answer 1

X + H2SO4

Zn + H2SO4 --------> ZnSO4 + H2

2M + 3H2SO4 ----------> M2(SO4)3 + 3H2

\(n_{H_2}=\dfrac{6,16}{22,4}=0,275\left(mol\right)\)

Ta có : \(M_X< \dfrac{7,3}{\dfrac{0,275.2}{3}}=39,82\) 

Mà Zn > 39,82

=> M phải < 39,82 (1)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(2M+6HCl\rightarrow2MCl_3+3H_2\)

\(\dfrac{1}{6}\)<------------------------------0,25

Vì \(n_{H_2}< 0,25\left(mol\right)\)

=> \(n_M< \dfrac{1}{6}\)

=> \(M_M>\dfrac{4,05}{\dfrac{1}{6}}=24,3\) ( 2)

Từ (1), (2) => M là Al