Thêm từ từ đến hết 100 ml dung dịch X

Chọn đáp án A

Giải thích:

$\begin{array}{l}+n_{HC{O_3}^{-}}=\mathrm{0,2};n_{C{O_3}^{2-}}=\mathrm{0,3}\Rightarrow \frac{n_{HC{O_3}^{-}}}{n_{C{O_3}^{2-}}}=\frac{2}{3}\Rightarrow \frac{n_{HC{O_3}^{-}pö}}{n_{C{O_3}^{2-}pö}}=\frac{2x}{3x}.\\ +n_{H^{+}}=n_{HCl}+2n_{H_{2}S{O}_4}=\mathrm{0,6};n_{S{O_4}^{2-}}=n_{H_{2}S{O}_4}=\mathrm{0,15.}\\ +ChotutuXvaoYsexayraphanungdongthoi\\ 2H^{+}+C{O}_3^{2-}\stackrel{}{\to }C{O}_2+H_{2}O\\ H^{+}+HC{O}_3^{-}\stackrel{}{\to }C{O}_2+H_{2}O\end{array}$

$\begin{array}{l}\Rightarrow n_{HC{O_3}^{-}pu}+2n_{C{O_3}^{2-}pu}=n_{H^{+}}\Rightarrow 2x+2.3x=\mathrm{0,6}\Rightarrow x=\mathrm{0,075.}\\ +\underset{ddZ}{\underbrace{\left\{\begin{array}{l}C{O_3}^{2-}:\mathrm{0,3}-\mathrm{0,075.3}=\mathrm{0,075}\\ S{O_4}^{-}:\mathrm{0,15}\\ HC{O_3}^{-}:\mathrm{0,05}\\ N{a}^{+},K^{+},C{l}^{-}\end{array}\right\}}}\stackrel{Ba{\left(OH\right)}_{2}du}{\to }\left\{\begin{array}{l}BaC{O}_3\downarrow :\mathrm{0,125}\\ BaS{O}_4\downarrow :\mathrm{0,15}\end{array}\right\}\\ \Rightarrow m_{kettua}=\mathrm{59,575}\end{array}$