Hỗn hợp X gồm Ba, BaO, Na, Na2O và K. C...

A

- Hướng dẫn giải

\(m(gam)X\left\{ \begin{array}{l}
Ba\\
Na\\
K\\
O
\end{array} \right. \to \left\langle \begin{array}{l}
{H_2}:0,14(mol)\\
Y\left\{ \begin{array}{l}
Ba{(OH)_2}:0,93m(g)\\
NaOH:0,18(mol)\\
KOH:0,044m(g)
\end{array} \right. \to 
\end{array} \right.\)

\(\begin{array}{l}
BTKL:{m_X} + {m_{{H_2}O}} = {m_Y} + {m_{{H_2}}}\\
 \to {m_{{H_2}O}} = 0,93m + 7,2 + 0,044m + 0,14.2 - m = 7,48 - 0,026m
\end{array}\)

\( \to {n_{{H_2}O}} = \frac{{7,48 - 0,026m}}{{18}}(mol)\)

\(BTNT(H):\frac{{7,48 - 0,026m}}{{18}} = \frac{{0,93m}}{{171}} + \frac{1}{2}.0,18 + \frac{1}{2}.\frac{{0,044m}}{{56}} + 0,14\)

\(\begin{array}{l}
 \to m = 25,5(gam)\\
 \to {n_{O{H^ - }(Y)}} = 2{n_{Ba{{(OH)}_2}}} + {n_{NaOH}} + {n_{KOH}} = 0,4774(mol)
\end{array}\)

\(O{H^ - } \to  \downarrow \)

\(T = \frac{{{n_{O{H^ - }}}}}{{{n_{C{O_2}}}}} = 1,37 \to \left\{ \begin{array}{l}
CO_3^{2 - }:a\\
HCO_3^ - :b
\end{array} \right. \to \left\{ \begin{array}{l}
a + b = 0,348\\
2a + b = 0,4774
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,1294\\
b = 0,2186
\end{array} \right.\)

\(\begin{array}{l}
B{a^{2 + }} + CO_3^{2 - } \to BaC{O_3}\\
{n_{B{a^{2 + }}}} = 0,1387 < {n_{CO_3^{2 - }}} = 0,1294\\
 \to {n_{BaC{O_3}}} = 0,1294(mol)\\
 \to m = 25,5(gam)
\end{array}\)