Cho 50 ml dung dịch AgNO3 vào 100 g dung dịch 1,9% muối

Đề bài

Cho 50 ml dung dịch AgNO ₃ vào 100 g dung dịch 1,9% muối chloride của một kim loại M hóa trị II, phản ứng vừa đủ thu được 5,74 g kết tủa AgCl. Biết PTHH của phản ứng là:

${\rm{MC}}{{\rm{l}}_{\rm{2}}}\,\,{\rm{ + }}\,\,{\rm{2AgN}}{{\rm{O}}_{\rm{3}}}\,\, \to \,\,{\rm{M(N}}{{\rm{O}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\,\,{\rm{ + }}\,\,\,{\rm{2AgCl}} \downarrow$

a) Xác định kim loại M.

b) Xác định nồng độ mol dung dịch AgNO ₃ .

a) Gọi số mol muối MCl ₂ là a.

Khối lượng muối: $\frac{{100\,{\rm{.}}\,\,{\rm{1,9\% }}}}{{{\rm{100\% }}}}{\rm{ = 1,9}}\,{\rm{(g)}}{\rm{.}}$

$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{MC}}{{\rm{l}}_{\rm{2}}}\,\,{\rm{ + }}\,\,{\rm{2AgN}}{{\rm{O}}_{{\rm{3}}\,}} \to \,M{{\rm{(N}}{{\rm{O}}_{\rm{3}}}{\rm{)}}_{\rm{2}}}\,\,{\rm{ + }}\,{\rm{2AgCl}}\\{\rm{Theo PTHH:}}\,\,\,\,\,\,\,\,{\rm{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(mol)}}\\{\rm{Phan ung :}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{a}}\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{2a}}\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,{\rm{a}}\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,{\rm{2a}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(mol)}}\,\,\,\,\,\,\,\,\,\,\\{\rm{Ta co: }}\,\,\,\,\,\,{\rm{a}}\,{\rm{ = }}\,\,\frac{{5,74}}{{{\rm{2}}\,{\rm{.}}\,{\rm{143,5}}}}\,{\rm{ = }}\,{\rm{0,02}}\,{\rm{(mol)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(1)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(M}}\,\,{\rm{ + }}\,\,{\rm{2}}{\rm{. 35,5)}}{\rm{.0,01}}\,\,{\rm{ = }}\,{\rm{ 0,95}}\,\,{\rm{ = > }}\,{\rm{M}}\,{\rm{ = }}\,{\rm{24}}\,\end{array}$

Vậy kim loại cần tìm là Mg.

b) Nồng độ của dung dịch AgNO ₃ : $\frac{{2\,{\rm{.}}\,\,0,02}}{{0,05}}{\rm{ = }}\,{\rm{0,8}}\,{\rm{(M)}}{\rm{.}}$