A. 54,65 gam.
B. 46,60 gam.
C. 19,70 gam.
D. 66,30 gam.
A
\(\begin{array}{l}
{H^ + } + CO_3^{2 - } \to HCO_3^ - \\
0,1 \leftarrow 0,1 \to \,\,\,0,1
\end{array}\)
\(\begin{array}{l}
HCO_3^ - + {H^ + } \to C{O_2} + {H_2}O\\
0,2\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,0,2
\end{array}\)
\( \to Y\left\{ \begin{array}{l}
N{a^{2 + }}\\
HCO_3^ - \\
SO_4^{2 - }
\end{array} \right.\)
\(\begin{array}{l}
{n_{N{a^ + }}} = 0,1.2 + 0,2.1 = 0,4(mol)\\
{n_{HCO_3^ - }} = 0,1 + 0,2 - 0,2 = 0,1(mol)
\end{array}\)
\( \to {n_{SO_4^{2 - }}} = \frac{{0,4 - 0,1}}{2} = 0,15(mol)\)
\(Y + Ba{(OH)_2} \to \left\{ \begin{array}{l}
HCO_3^ - + O{H^ - } \to CO_3^{2 - } + {H_2}O\\
0,1\,\, \to \,\,\,\,\,\,\,0,1\\
B{a^{2 + }} + CO_3^{2 - } \to BaC{O_3}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\, \to \,\,\,\,\,\,\,0,1\\
B{a^{2 + }} + SO_4^{2 - } \to BaS{O_4}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\, \to \,\,\,\,\,0,15
\end{array} \right.\)
\( \to m = 0,1.197 + 0,15.233 = 54,65(gam)\)
→ Chọn đáp án A.
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