Hướng dẫn:

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$

${{X}_{2}}+{{H}_{2}}\to 2HX$ (1)

       0,15 0,3                        (mol)

$HX+AgN{{O}_{3}}\to AgX\downarrow +HN{{O}_{3}}$           (2)

0,3                   0,3                        (mol)

Theo phương trình (1) có: ${{n}_{HX}}=2.{{n}_{{{H}_{2}}}}=2.0,15=0,3\left( mol \right)$

Theo phương trình (2) có: ${{n}_{AgX}}={{n}_{HX}}=0,3\left( mol \right)$

$\Rightarrow {{M}_{AgX}}=\frac{{{m}_{AgX}}}{{{n}_{AgX}}}=\frac{43,05}{0,3}$

${{M}_{Ag}}+{{M}_{Cl}}=143,5\Rightarrow {{M}_{X}}=143,5-{{M}_{Ag}}$

$\Rightarrow {{M}_{X}}=143,5-108=35,5$

Halogen là Clo

Chọn đáp án A