Đáp án:
\(\begin{array}{l}
{m_{MgC{l_2}}} = 9,5g\\
{m_{ZnC{l_2}}} = 27,2g\\
m = 36,7g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Mg(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
24a + 65b = 15,4
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,1mol\\
{m_{MgC{l_2}}} = 0,1 \times 95 = 9,5g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,2mol\\
{m_{ZnC{l_2}}} = 0,2 \times 136 = 27,2g\\
m = {m_{MgC{l_2}}} + {m_{ZnC{l_2}}} = 9,5 + 27,2 = 36,7g
\end{array}\)