Đáp án:
\(\begin{array}{l}
b)\\
{m_{Mg}} = 2,88g\\
{m_{Zn}} = 6,5g\\
c){m_{{\rm{dd}}HCl}} = 354g\\
d)\\
C{\% _{MgC{l_2}}} = 3,14\% \\
C{\% _{ZnC{l_2}}} = 3,75\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,22mol
\end{array}\)
Gọi a và b lần lượt là số mol của Mg và Zn
\(\begin{array}{l}
\left\{ \begin{array}{l}
24a + 65b = 9,38\\
a + b = 0,22
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,12\\
b = 0,1
\end{array} \right.\\
\to {n_{Mg}} = 0,12mol\\
\to {n_{Zn}} = 0,1mol
\end{array}\)
\(\begin{array}{l}
b)\\
{m_{Mg}} = 2,88g\\
{m_{Zn}} = 6,5g
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{Zn}} = 0,44mol\\
\to {m_{HCl}} = 16,1g\\
\to {m_{{\rm{dd}}HCl}} = (\dfrac{{16,1}}{{5\% }} \times 100\% ) \times \dfrac{{110}}{{100}} = 354g
\end{array}\)
\(\begin{array}{l}
d)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,12mol \to {m_{MgC{l_2}}} = 11,4g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1mol \to {m_{ZnC{l_2}}} = 13,6g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 363g\\
\to C{\% _{MgC{l_2}}} = \dfrac{{11,4}}{{363}} \times 100\% = 3,14\% \\
\to C{\% _{ZnC{l_2}}} = \dfrac{{13,6}}{{363}} \times 100\% = 3,75\%
\end{array}\)
