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Câu hỏi: 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 gam kết tủa. Lọc tách kết tủa, cô cạn dung dịch thu được $\mathrm{m}$ gam muối clorua. Giá trị của m là
A. $63,8.$
B. $22,6.$
C. $26,6.$
D. $15,0$.
A. $63,8.$
B. $22,6.$
C. $26,6.$
D. $15,0$.
$24,4\operatorname{gam}\left\{ \begin{array}{*{35}{l}}
\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} \\
~{{\text{K}}_{2}}\text{C}{{\text{O}}_{3}} \\
\end{array}+\text{BaC}{{\text{l}}_{2}}\to \text{BaC}{{\text{O}}_{3}}:39,4 \text{gam}+\left\{ \begin{array}{*{35}{l}}
\text{NaCl} \\
\text{KCl} \\
\end{array} \right. \right.$
${{n}_{\text{BaC}{{\text{l}}_{\text{2}}}}}={{n}_{\text{BaC}{{\text{O}}_{\text{3}}}}}=0,2~\text{mol}$
BTKL: ${{m}_{\text{hh}}}+{{m}_{\text{BaC}{{\text{l}}_{2}}}}={{m}_{\text{BaC}{{\text{O}}_{3}}}}+{{m}_{\text{m}}}$
$\to {{m}_{\text{m}}}=24,4+0,2.208-39,4=26,6 (~\text{g})$
\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: ${{m}_{\text{hh}}}+{{m}_{\text{BaC}{{\text{l}}_{2}}}}={{m}_{\text{BaC}{{\text{O}}_{3}}}}+{{m}_{\text{m}}}$
$\to {{m}_{\text{m}}}=24,4+0,2.208-39,4=26,6 (~\text{g})$
Đáp án C.