Cho 24,4 gam hỗn hợp \(\mathrm{Na}_{2} \mathrm{CO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}\) tác dụng vừa đủ với dung dịch \(\text{BaC}{{\text{l}}_{2}}.\) Sau phản ứng thu được 39,4 g...

* Hướng dẫn giải

Đáp án C.

\(24,4\operatorname{gam}\left\{ \begin{array}{*{35}{l}} \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} \\ ~{{\text{K}}_{2}}\text{C}{{\text{O}}_{3}} \\ \end{array}+\text{BaC}{{\text{l}}_{2}}\to \text{BaC}{{\text{O}}_{3}}:39,4\,\text{gam}+\left\{ \begin{array}{*{35}{l}} \text{NaCl} \\ \text{KCl} \\ \end{array} \right. \right.\)

\({{n}_{\text{BaC}{{\text{l}}_{\text{2}}}}}={{n}_{\text{BaC}{{\text{O}}_{\text{3}}}}}=0,2~\text{mol}\)

BTKL: \({{m}_{\text{hh}}}+{{m}_{\text{BaC}{{\text{l}}_{2}}}}={{m}_{\text{BaC}{{\text{O}}_{3}}}}+{{m}_{\text{m}}}\)

\(\to {{m}_{\text{m}}}=24,4+0,2.208-39,4=26,6\,(~\text{g})\)