Question

Answer 1

\(2X+2HCl->2XCl+H_2\)

\(\dfrac{4,6}{M_X}\).....0,2

n X= \(\dfrac{4,6}{M_X}\) mol

n HCl = \(\dfrac{10\%.73}{36,5}=0,2\) mol

=> n X = n HCl => \(\dfrac{4,6}{Mx}=0,2=>Mx=23đvc\)

vậy X là kim loại Natri (Na)

Answer 2

Ta có: \(m_{HCl}=\dfrac{10\%.73}{100\%}=7,3\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

2X + 2HCl ---> 2XCl + H₂

Theo PT: \(n_X=n_{HCl}=0,2\left(mol\right)\)

=> \(M_X=\dfrac{4,6}{0,2}=23\left(g\right)\)

Vậy X là natri (Na)