Đáp án:
Câu 5:
a) Al: 69,23%; Mg: 30,77%
b) HCl: 1,55%; AlCl3: 7,55%; MgCl2: 2,7%
Câu 6:
a) Ba: 85,09%; Mg: 14,91%
b) BaCl2: 9,64%; MgCl2: 4,4%
Giải thích các bước giải:
Câu 5:
${n_{{H_2}}} = 0,4mol;{n_{HCl}} = \dfrac{{339,95.1,02}}{{36,5}}.0,1 = 0,95mol$
Do $2{n_{{H_2}}} < {n_{HCl}}$ ⇒ $HCl$ dư
gọi x, y là số mol $Al$ và $Mg$
Ta có hpt:
$\left\{ \begin{gathered}
27x + 24y = 7,8 \hfill \\
3x + 2y = 0,4.2(BTe) \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x = 0,2 \hfill \\
y = 0,1 \hfill \\
\end{gathered} \right.$
$\% {m_{Al}} = \dfrac{{0,2.27}}{{7,8}}.100\% \approx 69,23\% $
$ \Rightarrow \% {m_{Mg}} = 30,77\% $
b) $\begin{gathered} {n_{HCldu}} = 0,95 - 2{n_{{H_2}}} = 0,15mol \hfill \\ {m_{ddA}} = {m_{KL}} + {m_{ddHCl}} - {m_{{H_2}}} = 7,8 + 339,95.1,02 - 0,4.2 = 353,749g \hfill \\ \end{gathered} $
$\begin{gathered} C{\% _{HCl}} = \dfrac{{0,15.36,5}}{{353,749}}.100\% = 1,55\% \hfill \\ C{\% _{AlC{l_3}}} = \dfrac{{0,2.133,5}}{{353,749}}.100\% = 7,55\% \hfill \\ C{\% _{MgC{l_2}}} = \dfrac{{0,1.95}}{{353,749}}.100\% = 2,7\% \hfill \\ \end{gathered} $
Câu 6:
a) ${n_{{H_2}}} = 0,2mol$
Gọi x, y là số mol $Ba$ và $Mg$
Ta có hpt: $\left\{ \begin{gathered}
137x + 24y = 16,1 \hfill \\
2x + 2y = 0,2.2(BTe) \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x = 0,1 \hfill \\
y = 0,1 \hfill \\
\end{gathered} \right.$
$\begin{gathered} \% {m_{Ba}} = \dfrac{{0,1.137}}{{16,1}}.100\% = 85,09\% \hfill \\ \% {m_{Mg}} = 14,91\% \hfill \\ \end{gathered} $
b) ${m_{ddX}} = {m_{KL}} + {m_{HCl}} - {m_{{H_2}}} = 16,1 + 200 - 0,4 = 215,7g$
$\begin{gathered} C{\% _{BaC{l_2}}} = \dfrac{{0,1.208}}{{215,7}}.100\% = 9,64\% \hfill \\ C{\% _{MgC{l_2}}} = \dfrac{{0,1.95}}{{215,7}}.100\% = 4,4\% \hfill \\ \end{gathered} $
