Đáp án:

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Giải thích các bước giải:

Câu 2:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,3mol
\end{array}\)

\(\begin{array}{l}
a.\\
{n_{{H_2}}} = {n_{Mg}} = 0,3mol\\
 \to {V_{{H_2}}} = 6,72l\\
{n_{HCl}} = 2{n_{Mg}} = 0,6mol\\
 \to {m_{HCl}} = 21,9g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{21,9}}{{7,3\% }} \times 100\%  = 300g
\end{array}\)

\(\begin{array}{l}
b.\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,3mol\\
 \to {m_{MgC{l_2}}} = 28,5g\\
{m_{{\rm{dd}}}} = {m_{Mg}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 306,6g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{28,5}}{{306,6}} \times 100\%  = 9,3\% 
\end{array}\)

Câu 3:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,5mol
\end{array}\)

Gọi a và b lần lượt là số mol của Mg và Fe

\(\begin{array}{l}
\left\{ \begin{array}{l}
24a + 56b = 18,4\\
a + b = 0,5
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,3\\
b = 0,2
\end{array} \right.\\
 \to {n_{Mg}} = 0,3mol\\
 \to {n_{Fe}} = 0,2mol
\end{array}\)

\(\begin{array}{l}
a.\\
\% {m_{Mg}} = \dfrac{{0,3 \times 24}}{{18,4}} \times 100\%  = 39,13\% \\
\% {m_{Fe}} = \dfrac{{0,2 \times 56}}{{18,4}} \times 100\%  = 60,87\% 
\end{array}\)

\(\begin{array}{l}
b.\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{Fe}} = 1mol\\
 \to {m_{HCl}} = 36,5g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{36,5}}{{7,3\% }} \times 100\%  = 500g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,3mol\\
 \to {m_{MgC{l_2}}} = 28,5g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
 \to {m_{FeC{l_2}}} = 25,4g\\
{n_{{H_2}}} = {n_{Mg}} + {n_{Fe}} = 0,5mol \to {m_{{H_2}}} = 1g
\end{array}\)

\(\begin{array}{l}
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 517,4g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{28,5}}{{517,4}} \times 100\%  = 5,5\% \\
 \to C{\% _{FeC{l_2}}} = \dfrac{{25,4}}{{517,4}} \times 100\%  = 4,9\% 
\end{array}\)

\(\begin{array}{l}
c.\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
{n_{NaOH}} = 2{n_{MgC{l_2}}} + 2{n_{FeC{l_2}}} = 1mol\\
 \to {V_{NaOH}} = \dfrac{1}{2} = 0,5l
\end{array}\)