Phương trình hóa học:
Kim loại $+\,\,\,\,2\,HCl\,\,\,\,\,\,\to \,\,\,\,\,\,$muối +$\,\,\,\,\,\,{{H}_{2}}$
${{n}_{{{H}_{2}}}}=\dfrac{2,24}{22,4}=0,1\,\,\,\,\left( mol \right)$
$\to {{n}_{HCl}}=2{{n}_{{{H}_{2}}}}=2.0,1=0,2\,\,\left( mol \right)$
Bảo toàn khối lượng
${{m}_{KL}}+{{m}_{HCl}}={{m}_{muối}}+{{m}_{{{H}_{2}}}}$
$\to 4,2\,\,\,+\,\,\,0,2.36,5\,\,\,=\,\,\,{{m}_{muối}}+0,2.1$
$\to {{m}_{muối}}=11,3\,\,\left( g \right)$