Đáp án:
a) 0,01 mol
b) 3,1875 g
Giải thích các bước giải:
a) Gọi x là số mol $Ba$, $Na$
${n_{{H_2}}} = \dfrac{{0,336}}{{22,4}} = 0,015mol$
$\begin{gathered}
2Na + 2{H_2}O \to 2NaOH + {H_2} \hfill \\
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2} \hfill \\
\end{gathered} $
⇒ ${n_{{H_2}}} = \dfrac{x}{2} + x = \dfrac{3}{2}x = 0,015 \Rightarrow x = 0,01$
b) Dung dịch Y gồm $N{a^ + }(0,01{\text{ mol); B}}{{\text{a}}^{2 + }}\left( {0,01{\text{ mol}}} \right);O{H^ - }\left( {0,03{\text{ mol}}} \right)$
${n_{{H^ + }}} = 0,125.0,1.2 = 0,025mol;{n_{C{u^{2 + }}}} = 0,125.0,1 = 0,0125mol$
${n_{SO_4^{2 - }}} = 0,125.0,1 = 0,0125mol$
$\begin{gathered}
{H^ + } + O{H^ - } \to {H_2}O \hfill \\
C{u^{2 + }} + 2O{H^ - } \to Cu{(OH)_2} \hfill \\
B{a^{2 + }} + SO_4^{2 - } \to BaS{O_4} \hfill \\
\end{gathered} $
$\begin{gathered}
\Rightarrow {n_{Cu{{(OH)}_2}}} = \dfrac{{0,03 - 0,0125}}{2} = 8,{75.10^{ - 3}}mol \hfill \\
{n_{BaS{O_4}}} = {n_{B{a^{2 + }}}} = 0,01mol \hfill \\
\Rightarrow {m_Z} = {m_{Cu{{(OH)}_2}}} + {m_{BaS{O_4}}} = 8,{75.10^{ - 3}}.98 + 0,01.233 = 3,1875\left( g \right) \hfill \\
\end{gathered} $
