Hòa tan m gam hỗn hợp chứa Zn và Al vào 500ml dung...

A

- Hướng dẫn giải

Số mol các chất là: \(\left\{ \begin{array}{l}
{n_{HCl}} = 0,5.0,4 = 0,2{\rm{ mol}}\\
{{\rm{n}}_{{H_2}S{O_4}}} = 0,5.0,3 = 0,15{\rm{ mol}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{n_{{H^ + }}} = 0,2 + 2.0,15 = 0,5{\rm{ mol}}\\
{{\rm{n}}_{C{l^ - }}} = 0,2{\rm{ mol}}\\
{n_{SO_4^{2 - }}} = 0,15{\rm{ mol}}
\end{array} \right.\) 

Sơ đồ phản ứng:

\(\left\{ \begin{array}{l}
Zn\\
Al
\end{array} \right\} + \left\{ \begin{array}{l}
\underbrace {HCl}_{0,2{\rm{ mol}}}\\
\underbrace {{H_2}S{O_4}}_{0,15{\rm{ mol}}}
\end{array} \right\} \to \left\langle \begin{array}{l}
{H_2} \uparrow \\
\underbrace {\left\{ \begin{array}{l}
Z{n^{2 + }},A{l^{3 + }},{H^ + }du\\
\underbrace {C{l^ - }}_{0,2{\rm{ mol}}},\underbrace {SO_4^{2 - }}_{0,15{\rm{ mol}}}
\end{array} \right\}}_{dd{\rm{ Y}}} \to \underbrace {\left\{ \begin{array}{l}
Zn{(OH)_2}\\
Al{(OH)_3}
\end{array} \right\} \downarrow }_{\max } + \underbrace {\left\{ \begin{array}{l}
N{a^ + }\\
\underbrace {C{l^ - }}_{0,2{\rm{ mol}}},\underbrace {SO_4^{2 - }}_{0,15{\rm{ mol}}}
\end{array} \right\}}_{dd{\rm{ sau}}}
\end{array} \right.\) 

\(\begin{array}{l}
 \to 1.{n_{N{a^ + }}} = 1.{n_{C{l^ - }}} + 2.{n_{SO_4^{2 - }}} \Rightarrow 1.{n_{N{a^ + }}} = 1.0,2 + 2.0,15\\
 \Rightarrow {n_{N{a^ + }}} = 0,5{\rm{ mol}}
\end{array}\) 

\( \to {n_{NaOH}} = {n_{N{a^ + }}} = 0,5{\rm{ mol}}\) 

\(V = {V_{dd{\rm{ NaOH}}}} = \frac{{{n_{NaOH}}}}{{{C_{M.NaOH}}}} = \frac{{0,5}}{1} = 0,5{\rm{ lit}}\) 

Đáp án A