Nhỏ từ từ dung dịch H2SO4 lo...

A

- Hướng dẫn giải

\(\begin{array}{l}
{H^ + } + CO_3^{2 - } \to HCO_3^ - \\
0,1 \leftarrow 0,1 \to \,\,\,0,1
\end{array}\) 

\(\begin{array}{l}
HCO_3^ -  + {H^ + } \to C{O_2} + {H_2}O\\
0,2\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,0,2
\end{array}\)             

\( \to Y\left\{ \begin{array}{l}
N{a^{2 + }}\\
HCO_3^ - \\
SO_4^{2 - }
\end{array} \right.\)

\(\begin{array}{l}
{n_{N{a^ + }}} = 0,1.2 + 0,2.1 = 0,4(mol)\\
{n_{HCO_3^ - }} = 0,1 + 0,2 - 0,2 = 0,1(mol)
\end{array}\)

\( \to {n_{SO_4^{2 - }}} = \frac{{0,4 - 0,1}}{2} = 0,15(mol)\)

\(Y + Ba{(OH)_2} \to \left\{ \begin{array}{l}
HCO_3^ -  + O{H^ - } \to CO_3^{2 - } + {H_2}O\\
0,1\,\, \to \,\,\,\,\,\,\,0,1\\
B{a^{2 + }} + CO_3^{2 - } \to BaC{O_3}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\, \to \,\,\,\,\,\,\,0,1\\
B{a^{2 + }} + SO_4^{2 - } \to BaS{O_4}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\, \to \,\,\,\,\,0,15
\end{array} \right.\)

\( \to m = 0,1.197 + 0,15.233 = 54,65(gam)\)

→ Chọn đáp án A.